**From: Matthew**

**Re: GAMSAT Chemistry Questions**

Below is a series of sample GAMSAT General Chemistry Questions which are just like the questions in the actual exam.

Our **GAMSAT Study Package** contains a complete practice test, which contains questions just like these.

Now, we’ll get started.

**GAMSAT General Chemistry Questions**

**(Do NOT use a calculator)**

Unit 1

Question 1-3

The thermodynamic properties of an ideal gas are described by the following equation of state:

P x V = n x R x T

where:

P is the pressure (atm)

V is the volume (L)

n is the amount of gas (mol)

R is the universal gas constant (0.082 L atm/K mol)

T is the temperature (K)

**1.** Which of the following values better approximate the volume of 0.1 moles of gas at P = 1 atm and room temperature (ca. 300 K)?

**A.** 0.3 L.

**B.** 3 L.

**C.** 30 L.

**D.** 300 L.

**2.** A gas is maintained at 200 K and 2 atm in a closed vessel. How many moles are kept in the small 10 mL container?

**A.** 10^{-1} mol.

**B.** 10^{-2} mol.

**C.** 10^{-3} mol.

**D.** 10^{-4} mol.

**3.** The volume of a gas (5 g), measured at 300 K and 3 atm, is 0.5 L. What is the molecular weight of this chemical species?

**A.** 10 g/mol.

**B.** 30 g/mol.

**C.** 100 g/mol.

**D.** 300 g/mol.

**Answers To GAMSAT General Chemistry Questions**

**Answer to Question 1**

From the above equation,

V = (n x R x T)/P = (0.1 mol x 0.082 L atm/K mol x 300 K)/ 1 atm ≌ 0.1 x 0.1 x 300 = 3 L

The key is therefore **B**.

**Answer to Question 2**

From the above equation,

n = (P x V)/R x T = (2 atm x 0.01 L)/ 0.082 L atm/K mol x 200 K ≌

≌ (2 x 0.01)/ (0.1 x 200) = 1 mmol

The key is therefore **C**.

**Answer to Question 3**

From the above equation,

n = (P x V)/R x T = (3 atm x 0.5 L)/ 0.082 L atm/K mol x 300 K ≌

≌ (3 x 0.5)/ (0.1 x 300) = 0.05 mmol

Also:

n = mass (g) / molecular weight (g/mol) = 5 g / MW = 0.05 mol

Hence:

MW = mass/n = 5 / 0.05 = 100 g/mol

The key is therefore **C**.

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