A solution of $1.25$ g of a certain non-volatile substance in $20$ g of water freezes at $271.94$ K. Calculate the molecular mass of the solute $(K_{f} = 1.86 K . k g . m o l^{- 1})$ .

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Solution

Freezing point of the solution $= 271.94 K$
Freezing point of water $= 273.0 K$

$△ T = (273 - 271.94) = 1.06 K$

We know that,

$m = \frac{1000 K_{f} \times w}{W \times △ T}$

Given,
$K_{f} = 1.86 K k g m o l^{- 1},$
$w = 1.25 g,$
$W = 20 g,$
$△ T = 1.06 K$

Now,

$M = \frac{1000 \times 1.86 \times 1.25}{20 \times 1.06}$

$M = 109.67 g / m o l$

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A solution of 1.25 g of a certain non-volatile substance in 20 g of water freezes at 271.94 K. Calculate the molecular mass of the solute (Kf=1.86 K.kg.mol−1).

Freezing point of the solution =271.94 K Freezing point of water =273.0 K △T=(273−271.94)=1.06KWe know that, m=1000Kf×wW×△TGiven, Kf=1.86 Kkgmol−1, w=1.25g,W=20g ,△T=1.06KNow,M=1000×1.86×1.2520×1.06M=109.67 g/mol

A solution of $1.25$ g of a certain non-volatile substance in $20$ g of water freezes at $271.94$ K. Calculate the molecular mass of the solute $(K_{f} = 1.86 K . k g . m o l^{- 1})$ .