If the molality of an aqueous solution is m a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For an aqueous solution,the solvent is water $(M_1 = 18 \ g/mol)$.Molality $m$ is defined as $m = \frac{n_2 	imes 1000}{W_1}$,where $n_2$ is mole

Vedclass · Accepted Answer

$y = \frac{18m}{1000+18m}$

Vedclass · Answer

(C) For an aqueous solution,the solvent is water $(M_1 = 18 \ g/mol)$.Molality $m$ is defined as $m = \frac{n_2 	imes 1000}{W_1}$,where $n_2$ is moles of solute and $W_1$ is mass of solvent in grams.Mole fraction of solute $y = \frac{n_2}{n_1 + n_2}$.Since $n_1 = \frac{W_1}{18}$,we have $W_1 = 18n_1$.Substituting $W_1$ in the molality formula: $m = \frac{n_2 	imes 1000}{18n_1} = \frac{1000}{18} 	imes \frac{n_2}{n_1}$.From the mole fraction definition,$\frac{n_2}{n_1} = \frac{y}{1-y}$.Therefore,$m = \frac{1000}{18} 	imes \frac{y}{1-y}$.Rearranging for $y$: $18m(1-y) = 1000y \implies 18m - 18my = 1000y \implies 18m = y(1000 + 18m)$.Thus,$y = \frac{18m}{1000 + 18m}$.

If the molality of an aqueous solution is $m$ and the mole fraction of the solute is $y$,which of the following relations is correct?

Similar Questions

If the density of a $1 \, m$ aqueous solution of $NaCl$ is $1.21 \, g/mL$,then the molarity of the solution is .......... (in $, M$)

The sum of the mole fraction of the components of a solution is

Give the meaning of $10\%$ glucose by weight.

The density of $2 \ M \ MgBr_2$ solution is $1.2 \ g/mL$. Then the molality of the solution is ........ $m$.

How many grams of $KCl$ would have to be dissolved in $60 \ g$ of $H_2O$ to give a $20 \%$ by weight solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module