Question

Consider two solutions, one formed by adding 10 g of glucose (C6H12O6) to 1 L of water and the other formed by adding 10 g of sucrose (C12H22O11) to 1 L of water.1) Calculate the vapor pressure for the first solution at 20 ?C. (The vapor pressure of pure water at this temperature is 17.5 torr.).2) Calculate the vapor pressure for the second solution at 20 ?C. (The vapor pressure of pure water at this temperature is 17.5 torr.).

Answer 1

Answer:

The vapor pressure for the first solution at 20°C is 17.48 Torr.

The vapor pressure for the second solution at 20°C is 17.49 Torr.

Explanation:

1) mass of solute that is glucose = 10 g

Moles of glucose  ,$n_1=\frac{10 g}{180 g/mol}=0.05555 mol$

Mass of water = m

Volume of water = V = 1L = 1000 mL

Density of water = 1 g/ml

$m=d\times V=1 g/mL\times 1000 mL=1000 g$

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{0.05555 mol}{0.05555 mol+55.55 mol}$

$p=17.48 Torr$

The vapor pressure for the first solution at 20°C is 17.48 Torr.

2 ) mass of solute that is sucrose= 10 g

Moles of sucrose ,$n_1=\frac{10 g}{342 g/mol}=0.02924 mol$

Mass of water = m

Volume of water = V = 1L = 1000 mL

Density of water = 1 g/ml

$m=d\times V=1 g/mL\times 1000 mL=1000 g$

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{0.02924 mol}{0.02924 mol+55.55 mol}$

$p=17.49 Torr$

The vapor pressure for the second solution at 20°C is 17.49 Torr.