Question

A solution is prepared by dissolving 6.00 g of an unknown nonelectrolyte in enough water to make 1.00 L of solution. The osmotic pressure of this solution is 0.750 atm at 25.0°C. What is the molecular weight (g/mol) of the unknown solute? g

Answer 1

Answer: The molar mass of unknown solute is 195.44 g/mol.

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=icRT$

where,

$\pi$ = osmotic pressure of the solution = 0.750 atm

i = Van't hoff factor = 1 (for non-electrolytes)

c = concentration of solute = ?

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$0.750atm=1\times c\times 0.0820\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\c=0.0307mol/L$

The concentration of solute is 0.0307 mol/L

This means that, 0.0307 moles are present in 1 L of solution.

To calculate the molecular mass of solute, we use the equation:

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

Moles of solute = 0.0307 mol

Given mass of solute = 6g

Putting values in above equation, we get:

$0.0307mol=\frac{6g}{\text{Molar mass of solute}}\\\\\text{Molar mass of solute}=195.44g/mol$

Hence, the molar mass of unknown solute is 195.44 g/mol.