Question

What is the equilibrium partial pressure of water vapor above a mixture of 62.9 g H2O and 33.2 g HOCH2CH2OH at 55 °C. The partial pressure of pure water at 55.0 °C is 118.0 mm Hg. Assume ideal behavior for the solution.

Answer 1

Answer : The partial pressure of $H_2O$ is 102.3 mmHg.

Explanation :

As per question,

Mass of $H_2O$ = 62.9 g

Mass of $HOCH_2CH_2OH$ = 33.2 g

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $HOCH_2CH_2OH$ = 62 g/mole

First we have to calculate the moles of $H_2O$ and $HOCH_2CH_2OH$.

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{62.9g}{18g/mole}=3.49mole$

$\text{Moles of }HOCH_2CH_2OH=\frac{\text{Mass of }HOCH_2CH_2OH}{\text{Molar mass of }HOCH_2CH_2OH}=\frac{33.2g}{62g/mole}=0.536mole$

Now we have to calculate the mole fraction of $H_2O$ and $HOCH_2CH_2OH$.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }HOCH_2CH_2OH}=\frac{3.49}{3.49+0.536}=0.867$

$\text{Mole fraction of }HOCH_2CH_2OH=\frac{\text{Moles of }HOCH_2CH_2OH}{\text{Moles of }H_2O+\text{Moles of }HOCH_2CH_2OH}=\frac{0.536}{3.49+0.536}=0.134$

Now we have to partial pressure of $H_2O$.

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of water vapor

$p_T$ = total pressure of gas

$X$ = mole fraction of water vapor

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.867\times 118.0mmHg=102.3mmHg$

Therefore, the partial pressure of $H_2O$ is 102.3 mmHg.