Question

A solution contains 3.5 gram of a non volatile solute in 125gm of water and it boils at 373.52K. calculate the molecular mass of solute (Kb for water =0.52)

Answer 1

Answer:

M=4.16 gm/mol

Explanation:

Using the formula for elevation in boiling point :

ΔT=Kb(m)----(1)

where Kb : ebullioscopic constant

           m : molality of solute

m= molality =no of moles of solute/ Kg of solvent

$\boxed{m=\frac{(3.5/M)}{0.125}}$

∴ Using the above and formula (1)

⇒ (373.52-370)=(0.52)($\frac{(3.5/M)}{0.125}$)

⇒ $M=\frac{0.52}{0.125}$

⇒ $\boxed{M=4.16 gm/mol}$

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