Question

A solution is prepared by dissolving 9.25g of non-volatile solute in 450 ml of water.it has osmotic pressure of 350 mm of hg at 27 c .assuming the solute in non-electrolyte,determine its molecular mass.

Answer 1

Molar mass of the solute is 24811.5 g/mol

Explanation:

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

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 350 mmHg

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

Mass of solute= 9.25 g   (Conversion factor: 1 g = 1000 mg)

Volume of solution = 450 mL

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

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

Putting values in above equation, we get:

$15.5mmHg=1\times \frac{9.25\times 1000}{\text{Molar mass of solute}\times 450}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 300K\\\\\text{molar mass of solute}=24811.5g/mol$

Hence, the molar mass of the solute is 24811.5 g/mol

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