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SAT II Chemistry : Solutions and States of Matter

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Example Questions

Example Question #1 : Gases

Suppose that an ideal gas at $0^{\circ}\text{C}$ $\displaystyle 0^{\circ}\text{C}$ and 2.0 atm has a volume of 11.2 cubic meters. Which of the following expressions represents the volume of the gas, in cubic meters, if the temperature were increased by $50^{\circ}\text{C}$ $\displaystyle 50^{\circ}\text{C}$ and the pressure decreased to 1.5 atm?

Possible Answers:

$11.2\text{ m}^3\times\frac{323\text{ K}}{273\text{ K}}\times\frac{1.5\text{ atm}}{2.0\text{ atm}}$ $\displaystyle 11.2\text{ m}^3\times\frac{323\text{ K}}{273\text{ K}}\times\frac{1.5\text{ atm}}{2.0\text{ atm}}$

$11.2\text{ m}^3\times\frac{50\text{ K}}{0\text{ K}}\times\frac{1.5\text{ atm}}{2.0\text{ atm}}$ $\displaystyle 11.2\text{ m}^3\times\frac{50\text{ K}}{0\text{ K}}\times\frac{1.5\text{ atm}}{2.0\text{ atm}}$

$11.2\text{ m}^3\times\frac{323\text{ K}}{273\text{ K}}\times\frac{2.0\text{ atm}}{1.5\text{ atm}}$ $\displaystyle 11.2\text{ m}^3\times\frac{323\text{ K}}{273\text{ K}}\times\frac{2.0\text{ atm}}{1.5\text{ atm}}$

$11.2\text{ m}^3\times\frac{273\text{ K}}{323\text{ K}}\times\frac{2.0\text{ atm}}{1.5\text{ atm}}$ $\displaystyle 11.2\text{ m}^3\times\frac{273\text{ K}}{323\text{ K}}\times\frac{2.0\text{ atm}}{1.5\text{ atm}}$

$11.2\text{ m}^3\times\frac{273\text{ K}}{323\text{ K}}\times\frac{1.5\text{ atm}}{2.0\text{ atm}}$ $\displaystyle 11.2\text{ m}^3\times\frac{273\text{ K}}{323\text{ K}}\times\frac{1.5\text{ atm}}{2.0\text{ atm}}$

Correct answer:

Explanation:

First of all, the unit degrees Celsius should be converted to Kelvins by adding 273. This means that the temperature increases from 273 K to 323 K.

Next, the combined gas law can be used.

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ $\displaystyle \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Plugging in the known values gives

$\frac{2.0\text{ atm}\times11.2\text{ m}^3}{273\text{ K}}=\frac{1.5\text{ atm}\times V_2}{323\text{ K}}$ $\displaystyle \frac{2.0\text{ atm}\times11.2\text{ m}^3}{273\text{ K}}=\frac{1.5\text{ atm}\times V_2}{323\text{ K}}$

and solving for the volume gives

$V_2=\frac{2.0\text{ atm}\times11.2\text{ m}^3\times323\text{ K}}{1.5\text{ atm}\times273\text{ K}}$ $\displaystyle V_2=\frac{2.0\text{ atm}\times11.2\text{ m}^3\times323\text{ K}}{1.5\text{ atm}\times273\text{ K}}$

Rearranging shows this is clearly equivalent to the correct answer, $11.2\text{ m}^3\times\frac{323\text{ K}}{273\text{ K}}\times\frac{2.0\text{ atm}}{1.5\text{ atm}}$ $\displaystyle 11.2\text{ m}^3\times\frac{323\text{ K}}{273\text{ K}}\times\frac{2.0\text{ atm}}{1.5\text{ atm}}$.

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Example Question #1 : Effusion

How much faster is the rate of effusion of $H_2 _{(g)}$ $\displaystyle H_2 _{(g)}$ than the rate of effusion of $H_2O_{(g)}$ $\displaystyle H_2O_{(g)}$?

Possible Answers:

$\displaystyle 9$

$\displaystyle 3$

$\frac{1}{3}$ $\displaystyle \frac{1}{3}$

$\displaystyle 18$

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

Correct answer:

$\displaystyle 3$

Explanation:

By Graham's Law, $\frac{Rate_1}{Rate_2}=\sqrt{\frac{M_2}{M_1}}$ $\displaystyle \frac{Rate_1}{Rate_2}=\sqrt{\frac{M_2}{M_1}}$. The molar mass of $H_2_{(g)}$ $\displaystyle H_2_{(g)}$ is 2 g/mol and the molar mass of $H_2O_{(g)}$ $\displaystyle H_2O_{(g)}$ is 18 g/mol. Thus, $\sqrt{\frac{18}{2}}=\sqrt{9}=3$ $\displaystyle \sqrt{\frac{18}{2}}=\sqrt{9}=3$

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Example Question #1 : Concentration And Units

30 mL of 1.0 M $HClO_{3}_{(aq)}$ $\displaystyle HClO_{3}_{(aq)}$ solution is diluted with water to a volume of 3 L. What is the new concentration of the solution?

Possible Answers:

$0.1\textup{\space M}$ $\displaystyle 0.1\textup{\space M}$

$10\textup{\space M}$ $\displaystyle 10\textup{\space M}$

$0.01\textup{\space M}$ $\displaystyle 0.01\textup{\space M}$

$0.3\textup{\space M}$ $\displaystyle 0.3\textup{\space M}$

$1\textup{\space M}$ $\displaystyle 1\textup{\space M}$

Correct answer:

$0.01\textup{\space M}$ $\displaystyle 0.01\textup{\space M}$

Explanation:

Recall that $\displaystyle 1 L = 1000 mL.$ Thus, the volume of the solution increased from 30 mL to 3000 mL, which is a factor of 100. Thus, the new concentration must be 100 times less than the original concentration, and $\frac{1 M}{100}=0.01 M$ $\displaystyle \frac{1 M}{100}=0.01 M$.

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SAT II Chemistry : Solutions and States of Matter

Study concepts, example questions & explanations for SAT II Chemistry

All SAT II Chemistry Resources

Example Questions

Example Question #1 : Gases

Example Question #1 : Effusion

Example Question #1 : Concentration And Units

All SAT II Chemistry Resources

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