Question

We have a methanol solution that is 6.0 m methanol, CH 3OH in water. Vapor pressure pure water at 30. degrees C is 31.82 mmHg. Vapor pressure methanol at 30. degrees C is 170.0 mm Hg. What is the total pressure above the solution at 30. degrees C

Answer 1

Answer: The total pressure above the solution at 30°C is 45.29 mmHg

Explanation:

We are given:

Molality of methanol = 6.0 m

This means that 6.0 moles of methanol present in 1 kg or 1000 g of pure water

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of water = 1000 g

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$\text{Moles of water}=\frac{1000g}{18g/mol}=55.56mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

$\chi_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}\\\\\chi_{CH_3OH}=\frac{6}{6+55.56}=0.0975$

$\chi_{water}=\frac{n_{water}}{n_{CH_3OH}+n_{water}}\\\\\chi_{water}=\frac{55.56}{6+55.56}=0.9025$

To calculate the total pressure, we use the equation given by Dalton and Raoults, which is:

$p_T=(p_A\times \chi_A)+(p_B\times \chi_B)$

where,

$p_T$ = total vapor pressure  = ?

We are given:

Mole fraction of methanol = 0.0975

Mole fraction of water = 0.9025

Vapor pressure of methanol = 170.0 mmHg

Vapor pressure of water = 31.82 mmHg

Putting values in above equation, we get:

$p_T=(170\times 0.0975)+(31.82\times 0.9025)\\\\p_T=45.29mmHg$

Hence, the total pressure above the solution at 30°C is 45.29 mmHg