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}=(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}=(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}=(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}=(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* 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)* p_T`

`p_(H_2O)=0.867* 118.0mmHg=102.3mmHg`

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