water contains weight solution|LIQUIDSOLUTIONJEEMain13-4-25

JEE MAIN 2023

13-4-2023 S2

Question

Sea water contains $29.25 \% \mathrm{NaCl}$ and $19 \% \mathrm{MgCl}_2$ by weight of solution. The normal boiling point of the sea water is $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$ (Nearest integer).
Assume $100 \%$ ionization for both NaCl and $\mathrm{MgCl}_2$
Given: $\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_2 \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Molar mass of NaCl and $\mathrm{MqCl}_2$ is 58.5 and $95 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively.

Write Your Answer

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Sol. Amount of solvent $=100-(29.25+19)=51.75 \mathrm{~g}$ $$ \begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=\left[\frac{2 \times 29.25 \times 1000}{58.5 \times 51.75}+\frac{3 \times 19 \times 1000}{95 \times 51.75}\right] \times 0.52 \\ & \Delta \mathrm{~Tb}=16.075 \\ & \Delta \mathrm{~Tb}=\left(\mathrm{T}_{\mathrm{b}}\right)_{\text {solution }}-\left(\mathrm{T}_{\mathrm{b}}\right)_{\text {solvent }} \\ & =\left(\mathrm{T}_{\mathrm{b}}\right)_{\text {solution }}=100+16.07=116.07^{\circ} \mathrm{C} \end{aligned} $$

Question Tags

JEE Main

Chemistry

Hard

Start Preparing for JEE with Competishun

Report Issue