## Learn key MCAT topics about MCAT math, plus practice questions and answers

(Note: This guide is part of our MCAT Physics series.)

### Part 2: Estimating and rounding numbers

a) Multiplication

b) Division

c) Example problem

### Part 3: Logarithms

a) Logarithm rules

b) Common applications of logarithms

c) Example problem

### Part 4: Strategies for eliminating answer choices

a) Significant figures

b) Exponents

c) Example problem

### Part 5: Fractions

b) Multiplication and division

c) Application: thin lens equation

d) Example problem

## Part 1: Introduction

While the MCAT does not permit the use of a calculator, the exam still contains questions that will require mathematical calculations—particularly in the context of physics and chemistry. Fine-tuning your math skills can thus significantly improve your ability to work through these test questions

While many of these math skills include basic addition, subtraction, multiplication, and addition, this section contains a few additional concepts that are important to brush up on for the MCAT, including logarithm rules and manipulating fractions. At the end of this guide, we have included several practice problems for you to practice with.

Let’s dive in!

## Part 2: Estimating and rounding numbers

Estimating and rounding numbers are useful skills on the MCAT, as they can help you calculate problems more effectively. This section contains important reminders to guide you through estimating and rounding.

### a) Multiplication

Suppose we’re trying to multiply 6.84 x 2.25. With no calculator available, how can we best approximate this answer in a short amount of time?

When multiplying numbers with many nonzero digits, rounding can give us more digestible numbers to work with. Rounding these numbers to one decimal place or a whole number can greatly simplify our mathematical calculations

To determine how much to round numbers, it may be helpful to first look at the answer choices that are presented. If the answer choices are closer together, then it would be better for us to round to one fewer digit to keep our calculations more accurate. If our answer choices are farther apart, then we can get away with rounding more.

To keep our calculations accurate, we can also compensate for our rounding for each digit. If we decide to round one number up, we should round the other number down, so that we stay closer to the true value. For instance, if we round 6.84 to 6.8, then we should round 2.25 to 2.3.

Suppose we chose to round 2.25 down to 2.2 instead. How would this compare to the correct answer and the answer provided by using 2.3?

$$6.84 \times 2.25 = 15.39$$   $$6.8 \times 2.3 = 15.64 \rightarrow 15.6$$   $$6.8 \times 2.2 = 14.96 \rightarrow 15.0$$

Our calculation is further from the correct answer when we use 2.2. By compensating for rounding and using the digit 2.3, we get a result closer to the actual answer of 15.39.

### b) Division

Estimating and rounding in division problems is similar to what we do for multiplication problems. As with multiplication, we first check the answer choices to determine how much we should round.

In division, however, we make different adjustments when we compensate for our rounding. To keep our numbers proportional to each other, both the dividend and divisor should be shifted in similar directions

Say we want to divide 19.58 by 4.67. If we decide to round 19.58 up to 20, then we should also round 4.67 up to 5. Doing so will give us a result of 4, which is close to the actual 4.19.

### c) Example problem

Let’s work through an example of a mathematical problem that is likely to be found on the MCAT

A student adds 6.19 grams of solid hydrochloric acid to 0.0500 L of water. What is the molarity of the resulting solution?

Molarity is given in moles per liter. To calculate the molarity of this solution, we will start by determining the number of moles of hydrochloric acid (HCl). First, we calculate the molar mass of HCl.

$$1.008 \space ^g/_{mol} \space + 35.45 \space ^g/_{mol} = 36.458 \space ^g/_{mol}$$

Then, we use this molar mass to convert the mass of HCl to moles. We can round both numbers down to simplify our calculation.

$$6.19 \space g \space HCl \div (36.458 \space ^g/_{mol} \space HCl)$$   $$6 \space g \space HCl \div (36 \space ^g/_{mol} \space HCl) = 0.1667 \space mol \space HCl$$

We then divide this value by the volume of the solution to determine molarity.

$$0.1667 \space mol \space HCl \space \div 0.0500 L = 3.33 M \space HCl$$

The approximate molarity of the solution is thus calculated to be 3.33M HCl.

## Part 3: Logarithms

### a) Logarithm rules

Logarithms are another mathematical concept likely to appear on the MCAT. These functions are the inverse of an exponential function, so they work by similar rules. Here are some important rules and operations to remember when manipulating logarithms:

• The log of 1 is always 0.

• The log of the base number is always 1.

• So long as the base numbers are the same, the log of two factors equals the sum of the log of the individual factors.

$$log_n(x\times y) = log_n(x) + log_n(y)$$

• Likewise, the log of a fraction equals the difference between the log of the dividend and the log of the divisor.

$$log_n(x \div y) = log_n(x) - log_n(y)$$

### b) Common applications of logarithms

One common logarithm you might see is with base “e.” These logarithms are known as natural logarithmsand are denoted as ln().

Logarithms with base 10 are also very common. (You should be very familiar with how to manipulate and approximate logarithms with base 10.) For the MCAT, it is important to note that the prefix p signifies –log. For instance, pH = -log[H+], and pOH = -log[OH-].

Decibels are another common application of logarithms in the MCAT. Decibels (dB) are a measure of sound level and can be calculated through the following equation:

$$dB = 10log(\frac{I}{I_0})$$    $$where \space I = \mbox{intensity of the sound, in } \frac{W}{m^2}$$   $$I_0 = \mbox{threshold of the lowest audible sound to a healthy human, equal to } 10^{-12} \frac{W}{m^2}$$

The Henderson-Hasselbalch equation is a common application of logarithms on the MCAT. This equation allows us to conduct buffer calculations and can be derived from the equilibrium constant expression for the dissociation of a weak acid. The Henderson-Hasselbalch equation is as follows:

$$pH = pK_a + log(\frac{[A^-]}{HA})$$

### c) Example problem

Let’s work through an application of using logarithms.

The pH of a solution of HF (pKa = 3.2) is 4.8. What is the ratio of acid to conjugate base in the HF solution?

The Henderson-Hasselbalch equation relates all of these values. We can use it to solve for the ratio.

$$pH = pK_a + log(\frac{[A^-]}{HA})$$

We start by isolating our ratio on one side of the equation

$$pH - pK_a = log(\frac{[A^-]}{[HA]})$$   $$-1 \times (pH - pK_a) = -1 \times log(\frac{[A^-]}{[HA]})$$   $$pK_a - pH = log(\frac{[HA]}{[A^-]})$$

From here, we can plug in our pKa and pH values.

$$3.2 - 4.8 = log(\frac{[HA]}{[A^-]})$$   $$-1.6 = log(\frac{[HA]}{[A^-]})$$

To obtain the ratio, we must apply the inverse of the logarithm functionthe exponential function. Since log() is the notation for base 10, we can raise the number 10 to the values present on both sides of the equation. Raising the number 10 to a logarithm with its own base will effectively “cancel” the logarithm function on that side of the equation

$$10^{-1.6} = \frac{[HA]}{[A^-]}$$

The ratio of HF to its conjugate base F - in this solution is thus calculated to be 10 -1.6 . We can further approximate this value by determining a range that this value should be on. While we do not know what the exact value of 10 -1.6 is, we do know the exact values of 10 -1 and 10 -2 :

$$10^{-2} \lt 10^{-1.6} \lt 10^{-1}$$   $$\frac{1}{100} \lt 10^{-1.6} \lt \frac{1}{10}$$   $$0.01 \lt 10^{-1.6} \lt 0.1$$

Thus, we would look for an answer option that lies somewhere between 0.01 and 0.1.

## Part 4: Strategies for eliminating answer choices

When working through calculation problems, it’s probably best to try calculating an exact answer, then checking to see if your answer matches an answer option. However, when in a time crunch, remember that you just need to pick 1 out of 4 answer choices! Below we will outline some tips to help improve your speed and accuracy in eliminating answer choices.

Scientific notation is a method of writing numbers with a significand and exponent.

• Significands are any real number whose absolute value is between 1 and 10, but not including 10.

• The exponent is base 10 and can be any whole number (i.e., negative, zero, or positive).

### a) Significant figures

Oftentimes, we can eliminate answer choices using significant figure calculations. When answer choices are written in scientific notation, focusing on the number of significant figures in the significand is useful.

Significant figures convey the certainty of a measurement based on its measuring instrument. They are necessary to consider in calculationsThe following rules can be used determine significant figures:

• All numbers between the first nonzero digit on the left and the last nonzero digit on the right are significant.

• When there is a decimal in the number, zeroes to the right of the last nonzero digit are significant. If there is no decimal, then those zeroes are not significant. (For instance, the number 4300.00 has six significant figures, but the number 4300 has two significant figures.)

• Leading zeroes, or zeros preceding the first nonzero digit, are not significant. (For instance, the number 0.010 has 2 significant figures.)

Conducting calculations with significant figures also requires specific rules.

• For calculations with addition and subtraction, begin by noting the position of the leftmost decimal point. From there, we proceed with calculating the problem and keep as many digits as we can until the very end of our calculation. After we have obtained this result, we round the calculated answer to the decimal place determined in the first step.

• For calculations with multiplication and division, first identify which factor, divisor, or dividend has the least number of significant digits. We then calculate the problem—again, keep as many digits until the very end. Finally, we round the calculated answer to the number of significant digits of the factor, divisor, or dividend determined in the first step.

### b) Exponents

Exponents are another method to eliminate answer choices. Because they are the inverse of the logarithm function, exponential functions follow similar rules:

$$A^0 = 1$$   $$A^x \times A^y = A^{x+y}$$   $$\frac{A^x}{A^y} = A^{x-y}$$   $$(A^x)^y = A^{x\times y}$$   $$(\frac{A}{B})^x = \frac{A^x}{B^x}$$   $$A^{-x} = \frac{1}{A^x}$$

When answer choices are written in scientific notation, the base number of the exponent is always 10. For questions that are addressing very small or very large quantities, we can check the exponent of this base number 10 to eliminate answer choices. For example, if a question asks for the number of molecules in a few moles of substance, we should expect a very large number. In this case, the answer will more likely contain an exponent raised to the 23rd power rather than the 5th power.

### c) Example problem

Let’s apply this knowledge with a practice problem.

Calculate the molarity of 0.424 mol HI in 2.1 L of aqueous solution

Let’s first identify which number has the fewest number of significant digits. The solution volume has fewer significant figures than the number of moles of HI. Because the volume has 2 significant figures, our final answer will also have 2 significant figures.

We can now calculate the problem, without rounding until the very end.

$$0.424 \mbox{ mol HI} \div 2.1 \space L = 0.201 \space M$$

We then rewrite this answer in scientific notation. First, we adjust the significand and multiply by the proper base ten exponent. Since we move the decimal over one place to the right, we multiply by 10 -1 .

$$0.2019 M \rightarrow 2.019 \times 10^{-1} \space M$$

We finally round this calculated answer to 2 significant figures to obtain our final answer.

$$0.2019 M \rightarrow 2.0 \times 10^{-1} \space M$$

When eliminating answer options, we can be sure to eliminate any answer options with large positive numbers in the exponent.

## Part 5: Fractions

The MCAT may require you to solve equations that deal with fractions. You may often see fractions applied in the context of the thin lens equation and solving optical systems.

When adding and subtracting fractions, we apply our operations to the numerators and keep the denominator. We must thus remember to find a common denominator before proceeding with our calculations. This is important in keeping a fraction balanced.

To pick a common denominator, we look at the least common multiple of the denominators we are working with. A least common multiple refers to the smallest number that both denominators are a factor of

For instance, if we want to add the following fractions:

$$\frac{2}{3} + \frac{1}{9}$$

We must find the smallest number that both 3 and 9 can be a factor of. In this case, the least common multiple is 9, since both 3 and 9 are factors of 9.

From here, we rewrite 2/3 as a fraction with a denominator of 9. To maintain the fraction’s proportionality, we must apply the same operation to the top and bottom. The denominator must be multiplied by 3 to produce the desired result of 9, so we multiply the top by 3 to get 6. After converting 2/3 to 6/9, we can proceed with adding.

$$\frac{6}{9} + \frac{1}{9} = \frac{6+1}{9} = \frac{7}{9}$$

### b) Multiplication and division

With multiplication, we multiply the numerators and denominators to obtain our result. In this case, we don’t need to find a common denominator before multiplying. Sometimes, however, we can simplify our calculations before multiplying. First, we check to see if either factor can be simplified in itself. Then, we check to see if the numerators of any fraction can be simplified with denominators of other fractions, or vice versa. Afterwards, we proceed with calculating.

For instance, if we’re multiplying the following fractions:

$$\frac{2}{7} \times \frac{14}{16}$$

We can start by simplifying 14/16 to 7/8. Our multiplication problem can then be rewritten as:

$$\frac{2}{7} \times \frac{7}{8}$$

From here, we can simplify with the numerators of one fraction that divides the denominator of another. Note that the number 7 divides both the denominator of the first fraction and the numerator of the second fraction

Additionally, the number 2 divides both the numerator of the first fraction and the denominator of the second fraction.

After simplifying these fractions, we are left with the following multiplication problem:

$$\frac{1}{1} \times \frac{1}{4}$$

In fraction multiplication, we multiply all values on the top half of the fraction together and multiply all values on the bottom half of the fraction together:

$$\frac{1}{1} \times \frac{1}{4} = \frac{1 \times 1}{1 \times 4} = \frac{1}{4}$$

The final answer should be ¼.

Division works in a similar fashion to multiplication. In cases of division, we multiply by the reciprocal of the divisor. For example, if we’re dividing the following fractions:

$$\frac{1}{3} \div \frac{4}{7}$$

Proceed by multiplying the first fraction by the reciprocal of the second fraction  (74). Then, we can proceed by the same rules as with multiplication:

$$\frac{1}{3} \div \frac{4}{7} = \frac{1}{3} \times \frac{7}{4} = \frac{1 \times 7}{3 \times 4} = \frac{7}{12}$$

The final answer should be 7/12.

### c) Application: thin lens equation

The thin lens equation is a common application of fraction math and physics on the MCAT. The thin lens equation relates different lengths associated with geometric optics: focal length, object distance, and image distance. Focal length (f) refers to the distance between the focal point and the center of a lens or mirror. Object length (o) refers to the distance between the object and the center of the lens or mirror. Image distance (i) is the distance between the image and the center of the lens or mirror.

Figure: The thin lens equation relates the distances f, i, and o.

The thin lens equation is as follows:

$$\frac{1}{f} = \frac{1}{o} + \frac{1}{i}$$   $$\mbox{where f = focal length,}$$   $$\mbox{o = object distance,}$$   $$\mbox{and i = image distance}$$

The sign of each variable contains information about its location

• Positive focal lengths are characteristic of concave mirrors and converging lenses; negative focal lengths are characteristic of convex mirrors and diverging lenses.

• Positive object distances occur when the object is in front of the mirror or lens, and negative object distances occur when the object is behind. In nearly all applications relevant to the MCAT, the object distance is positive.

• If the image distance is positive, the image is in front of the mirror or behind a lens. Such an image is referred to as a real image and is always inverted.

• If the image distance is negative, the image is behind the mirror or in front of a lens. Such an image is a virtual image and is always upright.

You can find more information about mirrors and lenses in our guide on light and optics .

When dealing with algebraic equations like the thin lens equation, remember to apply the same operations to both sides of the equation. For instance, if we subtract 4 from one side of the equation, we must also subtract 4 from the other side of the equation. Doing so keeps the equation balanced. Failing to do so may lead you to an incorrect answer!

### d) Example problem

Let’s work on an application of the thin lens equation

Suppose a converging lens with a focal length of 4 cm is placed 6 cm away from an object. Determine whether the image produced by this lens is real or virtual.

Start by writing down an equation you already know: the thin lens equation itself.

$$\frac{1}{f} = \frac{1}{o} + \frac{1}{i}$$

This question wants us to determine the value of the image distance, or i. We can isolate that variable on one side of the equation by subtracting 1/o from both sides of the equation

$$\frac{1}{i} = \frac{1}{f} - \frac{1}{o}$$

Now, we can evaluate the expression using our known values for f=4 cm and o=6 cm.

$$\frac{1}{i} = \frac{1}{4}\space cm^{-1} - \frac{1}{6}\space cm^{-1}$$

To subtract these numbers, we rewrite these fractions with a common denominator. In this example, we will use 4x6 = 24.

$$\frac{1}{i} = \frac{6}{24}\space cm^{-1} - \frac{4}{24}\space cm^{-1}$$

Continuing to evaluate the expression:

$$\frac{1}{i} = \frac{6-4}{24}\space cm^{-1}$$   $$\frac{1}{i} = \frac{2}{24}\space cm^{-1}$$

This brings us a bit closer to solving for i. To obtain an exact value for i, we will need to simplify these expressions without the use of fractions. Fortunately, we can cross-multiply here: multiply the numerator of the fraction on the left-hand side by the denominator of the fraction on the right-hand side, and vice versa.

$$1 \times 24 = 2 \times i$$   $$24 = 2i$$   $$24 \div 2 = i$$   $$i = 12 \space cm$$

Because our image distance is positive (i > 0), we know that the image is real.

Acknowledgements: Sneha Mittal

## Part 6: Passage-based questions and answers

In a college laboratory experiment, students were asked to complete a titration experiment with acetic acid as analyte and sodium hydroxide as titrant. To conduct the experiment, 100 mL of 0.200M NaOH was added to a burette. 50 mL of acetic acid of unknown concentration was added to an Erlenmeyer flask and placed beneath the burette. Students were asked to determine the concentration of the provided acetic acid (pKa = 4.7).

One of the students generates the following titration curve while completing the assignment experiment.

Figure 1: A student-generated titration curve.

Question 1: Approximately how many moles of NaOH were present in the burette at the equivalence point of the student’s titration?

A) 0.03 mol

B) 0.029 mol

C) 0.014 mol

D) 0.01 mol

Question 2: How many grams of NaOH were added to 100 mL water to create the solution used in the titration?

A) 0.8 g

B) 1.2 g

C) 0.5 g

D) 0.3 g

Question 3: What is the approximate ratio of [HA] to [A-] at the equivalence point of the titration?

A) 10 3.3

B) 10 4.3

C) 10 -4.3

D) 10 -3.3

Question 4: By how many orders of magnitude is the concentration of acetic acid after the student has added 14 mL NaOH greater than the concentration of acetic acid after the student has added 28 mL NaOH?

Question 5: Assume that instead of acetic acid, formic acid (pKa = 3.7) of the same concentration was used in this experiment. How would this affect the pH of the equivalence point?

A) The equivalence point would increase

B) The equivalence point would decrease

C) No change

D) Either A or C

1. Answer choice C is correct. At equivalence point, 28 mL NaOH was added to the Erlenmeyer flask, leaving 72 mL in the burette (100mL – 28mL = 72 mL). To determine the number of moles in 72 mL NaOH, we conduct the following calculations.

$$72 \space mL \space \times \frac{1 \space L}{1000 \space mL} \times 0.200 \space ^{mol}/_L =0.0144 \space mol$$

Given that one of the original numbers (72 mL) has 2 significant figures, we must also round our result to 2 significant figures. The correct answer is 0.014 mol.

2. Answer choice A is correct. Our calculations for this problem are similar to those from the last problem. In this case, we start with the total volume of NaOH mL and must also calculate the number of grams in one mole of NaOH.

Na has an atomic mass of 22.99 grams/mol, which we will round to 23 grams/mol. O has an atomic mass of 15.99 grams/mol, which we will round to 16 grams/mol. H has an atomic mass of 1.01 grams/mol, which we will round to 1 gram/mol.

Thus, the molar mass of NaOH is 23 + 16 + 1 = 40 g/mol.

At this stage, we can use dimensional analysis and use the known molarity of the solution to calculate the number of grams NaOH added:

$$100 \space mL \times \frac{1 \space L}{1000 \space mL} \times 0.200 \space ^{mol}/_L \times 40 \space ^g/_{mol} = 0.8 \space g$$

3. Answer choice C is correct. We can use the Henderson-Hasselbalch equation to solve for the ratio. At the equivalence point, the pH is approximately 9. The pKa of acetic acid is equal to 4.7. This question asks for us to find the ratio of [HA] to [A-], or [HA]/[A-].

$$pH = pK_a + log(\frac{[A^-]}{[HA]})$$   $$pH-pK_a = log(\frac{[A^-]}{[HA]})$$   $$-1 \times (pH - pK_a) = -1 \times log(\frac{[A^-]}{[HA]})$$   $$pK_a - pH = log(\frac{[HA]}{[A^-]})$$   $$4.7 - 9 = log(\frac{[HA]}{[A^-]})$$   $$-4.3 = log(\frac{[HA]}{[A^-]}$$   $$10^{-4.3} = \frac{[HA]}{[A^-]}$$

4. Answer choice D is correct. The pH around 14 mL NaOH is 2 and around 28 mL NaOH is 9. The difference between these numbers is 7, which represents the difference between the orders of magnitude of the two concentrations (choice D is correct). Answer C gives the exact ratio of these differences rather than the order of magnitude.

5. Answer choice B is correct. To answer this question, we must remember that pKa is the –log(Ka); thus, the higher the Ka value, the smaller the pKa value. Since the pKa of formic acid is less than that of acetic acid, its Ka value is higher. Because formic acid is a stronger acid, it will have a weaker conjugate base, which will not react as readily with water to produce formic acid and hydroxide. As a result of this decreased production of hydroxide, the pH of the solution will decrease (B).

## Part 7: Standalone questions and answers

Question 1: Suppose that a solution in a beaker is noted to have the following properties:

[HA] = 0.001, [A-] = 0.1 and pKa = 3.2

What is the approximate pH of this solution?

A) 10 -1.2

B) 10 -5.2

C) 1.2

D) 5.2

Question 2: How many significant figures are in 2090.10?

A) 4

B) 3

C) 5

D) 6

Question 3: What is the number 909.90 written in scientific notation?

A) 9.099 x 10 4

B) 9.099 x 10 2

C) 9.0990 x 10 2

D) 9.0990 x 10 4

Question 4: Sound A is 200 decibels and Sound B is 40 decibels. By how many orders of magnitude is Sound A greater than Sound B?

A) 160

B) 16

C) 5

D) 50

Question 5: A student is asked to prepare a 43.6 g sample of hypochlorous acid (HClO). How many moles of HClO should the student prepare?

A) 0.672 mol HOCl

B) 0.237 mol HOCl

C) 0.831 mol HOCl

D) 1.20 mol HOCl

### Answer key for standalone questions:

1. Answer choice D is correct. We can use the Henderson-Hasselbalch equation to solve for the pH. Three of the values used in the Henderson-Hasselbalch equation are given to us directly in the question stem.

$$pH = pK_a + log(\frac{[A^-]}{[HA]})$$   $$pH = 3.2 + log(\frac{0.1}{0.001})$$   $$pH = 3.2 + log(100)$$   $$pH = 3.2 + 2$$   $$pH = 5.2$$

2. Answer choice D is correct. Using the rules for significant figures, count all 3 digits to the left of the decimal point as significant. We also know to count the 2 zeroes in the hundreds and ones places as significant because they are between two nonzero numbers. The zero in the hundredths place is right of a nonzero number, making it significant as well. This amounts to 6 significant figures.

3. Answer choice C is correct. In scientific notation, the absolute value of the significand is less than 10. Thus, we rewrite 909.90 by moving the decimal point to the left two places. To compensate for this adjustment, we multiply the significand by a factor of 100. The only answer choice that does this and retains the number of significant figures from the original notation is C.

4. Answer choice B is correct. We can answer this question using the equation to calculate decibels. To compare two different sound levels, we can calculate the intensity of each sound, then calculate the ratio of both intensities.

Here, we will denote the intensity of sound A as I A  and the intensity of sound B as I B .   $$dB = 10 log(\frac{I}{I_0})$$   $$200 = 10 log(\frac{I_A}{I_0})$$   $$20 = log(\frac{I_A}{I_0})$$   $$10^{20} = \frac{I_A}{I_0}$$   $$I_0 \times 10^{20} = I_A$$   $$\mbox{Then, } 40 = 10 log(\frac{I_B}{I_0})$$   $$4 = log(\frac{I_B}{I_0})$$   $$10^4 = \frac{I_B}{I_0}$$   $$I_0 \times 10^4 = I_B$$   $$\mbox{Thus, } \frac{I_0 \times 10^{20}}{I_0 \times 10^4} = \frac{I_A}{I_B}$$   $$10^{16} = \frac{I_A}{I_B}$$

The ratio of I A / I B is 10 16 , indicating that Sound A is 16 orders of magnitude greater than Sound B.

5. Answer choice C is correct. To determine the number of moles, we convert this given mass using the molar mass of HOCl. First, we calculate the molar mass of HOCl.

The molar mass of H is 1.01 g/mol. We can round this to 1 g/mol.

The molar mass of O is 15.99 g/mol. We can round this to 16 g/mol.

The molar mass of HClO can thus be approximated as:

$$1 \space ^g/_{mol} \space H + 16 \space ^g/_{mol} \space O + 35.45 \space ^g/_{mol} \space Cl = 52.45 \space ^g/_{mol} \space HOCl$$

To calculate the number of moles in a sample with known mass, we must divide the mass by the molar mass.

$$43.6 \mbox{ g HOCl } \div (52.457 \space ^g/_{mol} \space HOCl)$$

At this point, we can eliminate answer option D. 43.6 is less than the number it is being divided by (52.47), so the answer to this division problem must be less than 1.

Since the answer choices are far apart from each other, we can round down our numbers to the nearest ten. This will give us a ratio that should be easier to calculate.

$$40 \space g \space HOCl \div (50 \space ^g/_{mol} \space HOCl) = 0.8 \mbox{ mol HOCl}$$

This estimated answer is fairly close to 0.831 mol, given by answer choice C.

What level of math is on the MCAT?
Any math that is on the MCAT is fundamental: just arithmetic, algebra, and trigonometry. There is absolutely no calculus on the MCAT. Math-based problems will appear mostly in the Chemical and Physical Foundations of Biological Systems section. more
Is math required for MCAT?
The MCAT is a conceptual exam that requires little mathematical computation. Any such requirements are based only on fundamental mathematical concepts, which include arithmetic, algebra, and trigonometry. There is no calculus involved. more
What math is on the MCAT?
What math is covered on the MCAT? The MCAT is primarily a conceptual exam, with little actual mathematical computation. Any math that is on the MCAT is fundamental: just arithmetic, algebra, and trigonometry. There is absolutely no calculus on the MCAT. more
How much math is in the MCAT?
What math is covered on the MCAT? The MCAT is primarily a conceptual exam, with little actual mathematical computation. Any math that is on the MCAT is fundamental: just arithmetic, algebra, and trigonometry. There is absolutely no calculus on the MCAT. more
Does MCAT have math?
What math is covered on the MCAT? The MCAT is primarily a conceptual exam, with little actual mathematical computation. Any math that is on the MCAT is fundamental: just arithmetic, algebra, and trigonometry. There is absolutely no calculus on the MCAT. more
Why is math on the MCAT?
While the MCAT does not permit the use of a calculator, the exam still contains questions that will require mathematical calculationsâ€”particularly in the context of physics and chemistry. Fine-tuning your math skills can thus significantly improve your ability to work through these test questions. more
Is the math on MCAT hard?
The good news: The math on the exam is not difficult and can be mastered with sufficient practice. This means that all calculations, regardless of the section of the MCAT, have to be completed mentally. more
How hard is math on MCAT?
The good news: The math on the exam is not difficult and can be mastered with sufficient practice. This means that all calculations, regardless of the section of the MCAT, have to be completed mentally. more
How important is math on the MCAT?
While the MCAT does not permit the use of a calculator, the exam still contains questions that will require mathematical calculationsâ€”particularly in the context of physics and chemistry. Fine-tuning your math skills can thus significantly improve your ability to work through these test questions. more
What kind of math does MCAT have?
Any math that is on the MCAT is fundamental: just arithmetic, algebra, and trigonometry. There is absolutely no calculus on the MCAT. Math-based problems will appear mostly in the Chemical and Physical Foundations of Biological Systems section. more
How hard is the math on MCAT?
The good news: The math on the exam is not difficult and can be mastered with sufficient practice. This means that all calculations, regardless of the section of the MCAT, have to be completed mentally. more

Source: www.shemmassianconsulting.com

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