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^\circ c 25.0 ∘
c. what is the molecular weight (g/mol) of the unknown solute?

Answer 1

Osmotic pressure can be calculated using the following equation:  

$\prod =C\times R\times T$

Here,

C representated concentration

R represented gas constant

T represented temperature

$\prod$  representated osmotic pressure

R=0.0821 atm L mol ⁻¹

T = 25 + 273 = 298 K

C=$\frac{Given mass of the substance}{Molar mass of the substance\times Volume}$  

Given mass of the substance= 6.00 g

Volume= 1 L

$\prod$ =0.75 atm

Putting all the values in the equation:

$0.750 =\frac{6}{Molar mass of the substance\times 1}\times 0.0821\times 298$

Molar mass of the substance= 195 g mol⁻¹.