Question: The freezing point of a solution prepared from $1.25gm$ of a non-electrolyte and $20gm$ of water...

Question

The freezing point of a solution prepared from $1.25gm$ of a non-electrolyte and $20gm$ of water is $271.9K$ . If molar depression constant is $1.86Kmole^{- 1}$ , then molar mass of the solute will be.

Answer 1

Solution

Molar mass $= \frac{K_{f} \times 1000 \times w}{\Delta T_{f} \times W} = \frac{1.86 \times 1000 \times 1.25}{20 \times 1.1}$

$= 105.68 = 105.7$ .

Question: The freezing point of a solution prepared from \(1.25gm\) of a non-electrolyte and \(20gm\) of water...

Solution