Find out the molality of the solution formed | vedclass

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

Question

(A) The molar mass of $NaCl$ is $23.0 + 35.5 = 58.5 \ g/mol$. Number of moles of $NaCl = \frac{	ext{mass}}{	ext{molar mass}} = \frac{29.22 \ g}{58.5

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(A) The molar mass of $NaCl$ is $23.0 + 35.5 = 58.5 \ g/mol$. Number of moles of $NaCl = \frac{	ext{mass}}{	ext{molar mass}} = \frac{29.22 \ g}{58.5 \ g/mol} = 0.4995 \ mol \approx 0.50 \ mol$. Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent. $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.50 \ mol}{2.00 \ kg} = 0.25 \ m$.

Find out the molality of the solution formed by dissolving $29.22 \ g$ of $NaCl$ in $2.00 \ kg$ of water. (in $m$)

Similar Questions

The normality of $4\% \ (w/V)$ $NaOH$ is

If the mole fraction of water in a solution of $H_2SO_4$ is $0.85$,then the molality of the solution is:

Calculate the mass of urea $(NH_2CONH_2)$ required in making $2.5 \, kg$ of $0.25 \, \text{molal}$ aqueous solution.

Calculate the mole fraction of $C_2H_6O_2$ (ethylene glycol) in an aqueous solution containing $28 \%$ by mass of $C_2H_6O_2$.

What is the mass of glucose in $500 \ g$ of an aqueous solution of glucose having a molality of $0.5 \ m$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module