$1.00$ g of non-electrolyte solute dissolved in $50$ g of benzene lowered the point of benzene by $0.40$ K. The freezing point depression constant of benzene is $5.12$ K kg/mol. Find the molar mass of the solute.

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Solution

As $Δ T_{f} = K_{f} \times m$

where,
$K_{f} =$ Freezing point depression constant;
$Δ T_{f} =$ Depression in freezing point;
$m =$ molality

Now,
$Δ T_{f} = K_{f} \times \frac{W_{2} \times 1000}{M_{2} \times W_{1}}$

where,
$W_{2} =$ mass of solute; $W_{1} =$ mass of benzene i.e. solvent; $M_{2} =$ molar mass of solute

$\Rightarrow 0.40 = 5.12 \times \frac{1.0 \times 1000}{M_{2} \times 50}$

$\Rightarrow M_{2} = 256$

Thus, the molar mass of non-electrolyte is $256$ .

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1.00g of non-electrolyte solute dissolved in 50 g of benzene lowered the point of benzene by 0.40K. The freezing point depression constant of benzene is 5.12 K kg/mol. Find the molar mass of the solute.

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