Respuesta :
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.
(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.
1. The volume of the ideal gas is [tex]\boxed{{\text{17}}{\text{.6 L}}}[/tex] .
2. The pressure of the ideal gas is [tex]\boxed{{\text{936 mm Hg}}}[/tex] .
3. The concentration of the solution, expressed as mass percent is [tex]\boxed{{\text{7}}{\text{.3 \% }}}[/tex] .
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:
[tex]{\text{PV}} = {\text{nRT}}[/tex] .......(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.
[tex]{\text{V}}=\dfrac{{{\text{nRT}}}}{{\text{P}}}[/tex] ......(2)
The pressure of the ideal gas is 950 mm Hg.
The temperature of the ideal gas is [tex]{\text{25}}\;^\circ{\text{C}}[/tex] .
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).
[tex]\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)}}\\&= 17.624{\text{ L}}\\&\approx {\text{17}}{\text{.6 L}}\\\end{aligned}[/tex]
Therefore the volume of the ideal gas is 17.6 L.
2. Rearrange equation (1) to calculate the pressure of ideal gas.
[tex]{\text{P}} =\dfrac{{{\text{nRT}}}}{{\text{V}}}[/tex] ......(3)
The volume of the ideal gas is 50 L.
The temperature of the ideal gas is [tex]{\text{27}}\;^\circ {\text{C}}[/tex] .
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).
[tex]\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}}}}\\&= 1.2321{\text{ atm}}\\&\approx 1.232{\text{ atm}}\\\end{aligned}[/tex]
The pressure is to be converted into mm Hg. The conversion factor for this is,
[tex]{\text{1 atm}} = {\text{760 mm Hg}}[/tex]
So the pressure of ideal gas can be calculated as follows:
[tex]\begin{aligned}{\text{P}} &= \left({{\text{1}}{\text{.232 atm}}}\right)\left( {\frac{{{\text{760 mm Hg}}}}{{{\text{1 atm}}}}}\right)\\&= 936.3{\text{2 mm Hg}} \\&\approx 93{\text{6 mm Hg}}\\\end{aligned}[/tex]
Therefore the pressure of the ideal gas is 936 mm Hg.
3. The formula to calculate the mass percent of KBr is as follows:
[tex]{\text{Mass}}\;{\text{percent}}=\left( {\dfrac{{{\text{Mass of KBr}}}}{{{\text{Mass of solution}}}}}\right)\left( {100} \right)[/tex] ......(4)
The mass of KBr is 12.4 g.
The mass of the solution is 170 g.
Substitute these values in equation (4).
[tex]\begin{aligned}{\text{Mass}}\;{\text{percent}}&= \left( {\frac{{{\text{12}}{\text{.4 g}}}}{{{\text{170 g}}}}} \right)\left( {100} \right)\\&= 7.294\;\% \\&\approx 7.{\text{3 \% }}\\\end{aligned}[/tex]
Therefore the concentration of the solution is 7.3 %.
Learn more:
1. Which statement is true for Boyle’s law: https://brainly.com/question/1158880
2. Calculation of volume of gas: https://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
