Question

A 0.500-g sample of a compound is dissolved in enough water to form 100.0 mL of solution. This solution has an osmotic pressure of 2.50 atm at 25C. If each molecule of the solute dissociates into two particles (in this solvent), what is the molar mass of this solute?

Short answer

The molar mass of this solute is 1 g

Step-by-step solution

Step 1:

The osmotic pressure of a solution is given by the equation

$\pi =i\times M\times R\times T$

Where i = van’t Hoff factor

M = molarity of the solution

R = ideal gas constant =$0.082atmLmo{l}^{-1}{K}^{-1}$

T = absolute temperature (note that temperature should be always

taken in kelvin)

i = 2 [ since solute dissociates into 2 particles]

$\pi =2.5atm$

R =$0.082atmLmo{l}^{-1}{K}^{-1}$

T= (273+25) K= 298K

Putting the given values in the above equation we get

$M=\frac{\pi }{i\times R\times T}=\frac{2.5}{2\times 0.082\times 298}=0.051M$

Step 2:

The molarity of the solution =$\frac{\text{molesofsolute}}{\text{volumeofsolutioninlitres}}$

The volume of solution in liters =$\frac{100}{1000}=0.1L$

0.051 M = $\frac{\text{molesofsolute}}{0.1L}$

Moles of solute = 0.050.1 = 0.5 mol

Moles of solute =$\frac{\text{givenmass}}{\text{molarmass}}$

Given mass of solute is 0.5 g

Molar mass =$\frac{0.5}{0.5}=1g$