1. Tardigrade
  2. Question
  3. Chemistry
  4. Dry air was passed successively through a solution of 5 g of a solute in 180 g of water and then through pure water. The loss in weight of solution was 2.50 g and that of pure solvent 0.04 g. The molecular weight of the solute is:

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Dry air was passed successively through a solution of 5 g of a solute in 180 g of water and then through pure water. The loss in weight of solution was 2.50 g and that of pure solvent 0.04 g. The molecular weight of the solute is:

Solutions

Solution:

$P_{0}-P_{s} \propto$ loss in weight of water chamber and $P_{s} \propto$ loss in wt. of solution chamber
$\frac{P^{o}-P_{s}}{P^{o}}=\frac{n}{N}=\frac{w\times M}{m\times W}$
or $\frac{0.04}{2.50}=\frac{5\times18}{m\times180}$
$\therefore m=31.25$