Question

Current of dry air was passed through a solution of 2.5 gram of a non volatile solute in 100 gram of water and through water alone the loss in the weight of solution was 1.25 grams and that of water was 0.05 gram calculate the molecular mass of the solute​

Answer 1

Answer:

hi

Explanation:

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Answer 2

Answer:

The molar mass of solute is found equal to 112.5gmol⁻¹.

Explanation:

Given: mass of the solute  $m_{2}=2.5g$

Weight of the solvent(water) $m_{1}=100g$

Molar mass of solute, $M_{2}$

We know the molar mass of water $M_{1}=18gmol^{-1}$

$P_{o}-P_{s}$  ∝ weight loss in pure water

$P_{s}$  ∝ weight loss is solution

Now   $\frac{P_{o}-P_{s}}{P_{s}} =\frac{n_{2}}{n_{1}}=\frac{m_{2}M_{1}}{m_{1}M_{2}}$                                        ................(1)

Weight loss in solution $n_{1}=1.25g$

Weight loss in pure water $n_{2}=0.005g$

Put all these values in eq. (1)

$\frac{0.005}{1.25} =\frac{(2.5)(18)}{M_{2}*100}$

∴    $M_{2}=112.5gmol^{-1}$

Therefore the molar mass of given solute in the solution would be equal to 112.5g/mol.