Question

A solution is prepared by dissolving 4.78 g of an unknown nonelectrolyte in enough water to make 375 mL of solution. The osmotic pressure of the solution is 1.33 atm at 27C. What is the molar mass of the solute?

Answer 1

Answer: The molar mass of the solute is 236.0 g/mol

Step-by-step explanation:

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

`\pi=iMRT`

where,

`\pi` = osmotic pressure of the solution = 1.33 atm

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

M = molarity of solute = ?

R = Gas constant =
`0.0821\text{ L atm }mol^(-1)K^(-1)`

T = temperature of the solution =
`27^oC=[273+27]K=300K`

Putting values in above equation, we get:

`1.33atm=1* M* 0.0821\text{ L.atm }mol^(-1)K^(-1)* 300K\\\\M=(1.33)/(1* 0.0821* 300)=0.0540M`

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}`

We are given:

Molarity of solution = 0.0540 M

Given mass of solute = 4.78 g

Volume of solution = 375 mL

Putting values in above equation, we get:

`0.0540M=\frac{4.78* 1000}{\text{Molar mass of solute}* 375}\\\\\text{Molar mass of solute}=(4.78* 1000)/(0.0540* 375)=236.0g/mol`

Hence, the molar mass of the solute is 236.0 g/mol