Question

Determine the vapor pressure of a solution at 55°c that contains 34.2 g nacl in 375 ml of water. the vapor pressure of pure water at 55°c is 118.1 torr. the van't hoff factor for nacl is 1.9 determine the vapor pressure of a solution at 55°c that contains 34.2 g nacl in 375 ml of water. the vapor pressure of pure water at 55°c is 118.1 torr. the van't hoff factor for nacl is 1.9 115 torr 92.8 torr 108 torr 87.1 torr 112 torr

Answer 1

Answer is: pressure is 112 torr.
n(NaCl) = 34,2 g ÷ 58,5 g/mol.
n(NaCl) = 0,55 mol.
m(H₂O) = 369,37 g.
n(H₂O)  = 369,37 g ÷ 18 g/mol = 20,52 mol.
mol fraction(H₂O) = 20,52 mol ÷ 21,7 mol = 0,95.
p = mol fraction(H₂O) · p(H₂O).
p = 0,997 · 118,1 torr.
p = 112 torr.

Answer 2

The vapor pressure of a solution that contains 34.2 g of NaCl in 375 mL wateris $\boxed{112{\text{ torr}}}$.

Further Explanation:

Colligative properties are dependent on concentration of solute and not on their identities.

Below mentioned are colligative properties.

1. Relative lowering of vapor pressure

2. Elevation in boiling point

3. Depression in freezing point

4. Osmotic pressure

When non-volatile solute is added to any solution, there occurs a decrease in vapor pressure of solution. This decrease in vapor pressure is called relative lowering of vapor pressure.

The formula to calculate moles of component is as follows:

${\text{Moles of component}} = \dfrac{{{\text{Mass of component}}}}{{{\text{Molar mass of component}}}}$                                             …… (1)

Substitute 34.2 g for mass of component and 58.5 g/mol for molar mass of component in equation (1) to calculate moles of NaCl.

$\begin{aligned}{\text{Moles of NaCl}} &= \frac{{{\text{34}}{\text{.2 g}}}}{{{\text{58}}{\text{.5 g/mol}}}} \\ &= 0.585{\text{ mol}} \\\end{aligned}$  

Since density of water is 1 g/mL, mass of water becomes 375 g. Substitute 375 g for mass of component and 18 g/mol for molar mass of component in equation (1) to calculate moles of ${{\text{H}}_{\text{2}}}{\text{O}}$.

 $\begin{aligned}{\text{Moles of }}{{\text{H}}_{\text{2}}}{\text{O}} &= \frac{{{\text{375 g}}}}{{{\text{18 g/mol}}}} \\&= 20.83{\text{ mol}} \\\end{aligned}$

The formula to calculate mole fraction of ${{\text{H}}_{\text{2}}}{\text{O}}$ is as follows:

${x_{{{\text{H}}_{\text{2}}}{\text{O}}}} = \dfrac{{{\text{Moles of }}{{\text{H}}_{\text{2}}}{\text{O}}}}{{{\text{Moles of }}{{\text{H}}_{\text{2}}}{\text{O}} + \left( {\left( {{\text{Moles of NaCl}}} \right)\left( {\text{i}} \right)} \right)}}$                             …… (2)

Here, i is the van’t Hoff factor for NaCl.

Substitute 0.585 mol for moles of NaCl, 20.83 mol for moles of ${{\text{H}}_{\text{2}}}{\text{O}}$ and 1.9 for i in equation (2).

$\begin{aligned}{x_{{{\text{H}}_2}{\text{O}}}} &= \frac{{{\text{20}}{\text{.83 mol}}}}{{{\text{20}}{\text{.83 mol}} + \left( {\left( {{\text{0}}{\text{.585 mol}}} \right)\left( {{\text{1}}{\text{.9}}} \right)} \right)}} \\&= 0.949 \\\end{aligned}$  

The formula to calculate vapor pressure of ${{\text{H}}_{\text{2}}}{\text{O}}$ is as follows:

${P_{{{\text{H}}_2}{\text{O}}}} = {x_{{{\text{H}}_2}{\text{O}}}}P_{{{\text{H}}_2}{\text{O}}}^{\text{o}}$                                                                                    …… (3)

Where,

${P_{{{\text{H}}_2}{\text{O}}}}$ is the vapor pressure of water.

${x_{{{\text{H}}_2}{\text{O}}}}$ is the mole fraction of water.

$P_{{{\text{H}}_2}{\text{O}}}^{\text{o}}$ is the vapor pressure of pure water.

Substitute 0.949 for ${x_{{{\text{H}}_2}{\text{O}}}}$ and 118.1 torrfor $P_{{{\text{H}}_2}{\text{O}}}^{\text{o}}$ in equation (3).

$\begin{aligned}{P_{{{\text{H}}_2}{\text{O}}}} &= \left( {0.949} \right)\left( {118.1{\text{ torr}}} \right) \\&= 112.0769{\text{ torr}} \\&\approx {\text{112 torr}} \\\end{aligned}$  

Learn more:

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Colligative properties

Keywords: colligative properties, relative lowering of vapor pressure, mass, molar mass, NaCl, H2O, vapor pressure, pure water, xH2O, PH2O, mole fraction, 112 torr, 0.949.