1. What is the volume of 0.900 moles of an ideal gas at 25.0° C and a pressure of 950.0 mm Hg?1.50 liters
17.6 liters
18.0 liters
34.2 liters

2. What is the pressure, in mm Hg, of 2.50 moles of an ideal gas if it has a volume of 50.0 liters when the temperature is 27.0° C?
84.2 mm Hg
289 mm Hg
617 mm Hg
936 mm Hg

3. When 12.4 grams of KBr are dissolved in enough water to create a 170-gram solution, what is the solution's concentration, expressed as a percent by mass?
6.2% KBr
6.8% KBr
7.3% KBr
7.9% KBr

Answers

Answer 1

Answer: 1. Using the ideal gas law, which states that PV = nRT, and remembering that absolute temperature must be used:
(950.0 mmHg)(V) = (0.900 moles)(R)(25.0 + 273.15 K)
The most convenient value of R is 0.08206 L-atm/mol-K, so we convert the pressure of 950.0 mmHg to atm by dividing by 760: 1.25 atm
(1.25 atm)(V) = (0.9)(0.08206)(298.15)
V = 17.6 L

2. Again using the ideal gas law, and converting temperature to Kelvin: 27 + 273.15 = 300.15 K
PV = nRT
P(50.0 L) = (2.50 moles)(0.08206 L-atm/mol-K)(300.15 K)
P = 1.23 atm
Then we multiply by 760 mmHg / 1 atm = 936 mmHg. This is the fourth of the choices.

3. If the complete solution has a mass of 170 grams, and 12.4 grams of it is the dissolved KBr, we simply have to divide the mass of solute (KBr) by the total mass of the solution: This gives us 12.4 grams / 170 grams = 0.0729. Multiplying by 100% gives 7.29% or approximately 7.3% KBr, which is the third of the choices.

Answer 2

Answer:

1. The volume of the ideal gas is $\boxed{{\text{17}}{\text{.6 L}}}$ .

2. The pressure of the ideal gas is $\boxed{{\text{936 mm Hg}}}$ .

3. The concentration of the solution, expressed as mass percent is $\boxed{{\text{7}}{\text{.3 \% }}}$ .

Further Explanation:

An ideal gas is a hypothetical gas that is composed of a large number of randomly moving particles that are supposed to have perfectly elastic collisions among themselves. It is just a theoretical concept and practically no such gas exists. But gases tend to behave almost ideally at a higher temperature and lower pressure.

Ideal gas law is the equation of state for any hypothetical gas. The expression for the ideal gas equation is as follows:

${\text{PV}} = {\text{nRT}}$ .......(1)

Here,

P is the pressure of ideal gas.

V is the volume of ideal gas.

T is the absolute temperature of the ideal gas.

n is the number of moles of the ideal gas.

R is the universal gas constant.

1. Rearrange equation (1) to calculate the volume of the ideal gas.

${\text{V}}=\frac{{{\text{nRT}}}}{{\text{P}}}$ ......(2)

The pressure of the ideal gas is 950 mm Hg.

The temperature of the ideal gas is ${\text{25}}\;^\circ{\text{C}}$ .

The number of moles of the ideal gas is 0.9 mol.

The universal gas constant is 0.0821 L atm/K mol.

Substitute these values in equation (2).

$\begin{aligned}{\text{V}}&=\frac{{\left( {{\text{0}}{\text{.9 mol}}} \right)\left( {0.0821{\text{ L atm/K mol}}} \right)\left( {25 + 27{\text{3}}{\text{.15}}}\right){\text{K}}}}{{\left( {950{\text{ mm Hg}}}\right)\left( {\frac{{{\text{1 atm}}}}{{760{\text{ mm Hg}}}}}\right)}}\n&= 17.624{\text{ L}}\n&\approx {\text{17}}{\text{.6 L}}\n\end{aligned}$

Therefore the volume of the ideal gas is 17.6 L.

2. Rearrange equation (1) to calculate the pressure of ideal gas.

${\text{P}} =\frac{{{\text{nRT}}}}{{\text{V}}}$ ......(3)

The volume of the ideal gas is 50 L.

The temperature of the ideal gas is ${\text{27}}\;^\circ {\text{C}}$ .

The number of moles of the ideal gas is 2.5 mol.

The universal gas constant is 0.0821 L atm/K mol

Substitute these values in equation (3).

$\begin{aligned}{\text{P}}&= \frac{{\left( {{\text{2}}{\text{.5 mol}}} \right)\left( {0.0821{\text{ L atm/K mol}}} \right)\left( {27 + 27{\text{3}}{\text{.15}}} \right){\text{K}}}}{{{\text{50 L}}}}\n&= 1.2321{\text{ atm}}\n&\approx 1.232{\text{ atm}}\n\end{aligned}$

The pressure is to be converted into mm Hg. The conversion factor for this is,

${\text{1 atm}} = {\text{760 mm Hg}}$

So the pressure of ideal gas can be calculated as follows:

$\begin{aligned}{\text{P}} &= \left({{\text{1}}{\text{.232 atm}}}\right)\left( {\frac{{{\text{760 mm Hg}}}}{{{\text{1 atm}}}}}\right)\n&= 936.3{\text{2 mm Hg}} \n&\approx 93{\text{6 mm Hg}}\n\end{aligned}$

Therefore the pressure of the ideal gas is 936 mm Hg.

3. The formula to calculate the mass percent of KBr is as follows:

${\text{Mass}}\;{\text{percent}}=\left( {\frac{{{\text{Mass of KBr}}}}{{{\text{Mass of solution}}}}}\right)\left( {100} \right)$ ......(4)

The mass of KBr is 12.4 g.

The mass of the solution is 170 g.

Substitute these values in equation (4).

$\begin{aligned}{\text{Mass}}\;{\text{percent}}&= \left( {\frac{{{\text{12}}{\text{.4 g}}}}{{{\text{170 g}}}}} \right)\left( {100} \right)\n&= 7.294\;\% \n&\approx 7.{\text{3 \% }}\n\end{aligned}$

Therefore the concentration of the solution is 7.3 %.

Learn more:

1. Which statement is true for Boyle’s law: brainly.com/question/1158880

2. Calculation of volume of gas: brainly.com/question/3636135

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole Concept

Keywords: P, V, n, R, T, ideal gas, pressure, volume, 17.6 L, 936 mm Hg, 7.3 %, 0.9 mol, 950 mm Hg, 50 L, 2.5 mol, 12.4 g, 170 g, KBr

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