LS_JEE-Main 12-4-19_S2 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN 2019

12-04-2019 S2

Question

A solution is prepared by dissolving 0.6 g of urea (molar mass $=60 \mathrm{~g} \mathrm{~mol}^{-1}$ ) and 1.8 g of glucose (molar mass $\left.=180 \mathrm{~g} \mathrm{~mol}^{-1}\right)$ in 100 mL of water at $27^{\circ} \mathrm{C}$. The osmotic pressure of the solution is $\left(\mathrm{R}=0.08206 \mathrm{~L} \mathrm{~atm} \mathrm{~K}_{-1} \mathrm{~mol}^{-1}\right)$

Select the correct option:

1.64 atm

2.46 atm

8.2 atm

4.92 atm

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Osmotic pressure $(\pi)=$ CRT
Solute: urea and glucose $$ \begin{aligned} & \therefore \pi=\left(C_1+C_2\right) R T \\ & =\left(\frac{0.6}{60 \times 0.1}+\frac{1.8}{180 \times 0.4}\right) \times 0.0821 \times 300 \\ & =0.2 \times 0.0821 \times 300 \\ & =4.926 \mathrm{~atm} \end{aligned} $$

Question Tags

JEE Main

Chemistry

Easy

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