Test Index

Solutions Part 2

Section: Chemistry

Question : 15 of 95

Marks: +1, -0

60\,\mathrm{g}\,\mathrm{mol}^{-1}

180\,\mathrm{g}\,\mathrm{mol}^{-1}

100\,\mathrm{mL}\,\mathrm{H}_2\mathrm{O}

27^{\circ}\,\mathrm{C}

R = 0.082\,\mathrm{L}\,\mathrm{atm}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}

Solution:

Concept:

Osmotic pressure is a colligative property that depends on the total number of solute particles in solution.

Explanation:

First, calculate moles of solute A:

n_A = \frac{0.3}{60} = 0.005

mol

Then, calculate moles of solute B:

n_B = \frac{0.9}{180} = 0.005

mol

Since both solutes are non‑electrolytes, they do not dissociate.

Total moles of solute particles:

n = n_A + n_B = 0.005 + 0.005 = 0.01

mol

Volume of solution (dilute, so approximately equal to water volume):

V = 100

= 0.1

Temperature in Kelvin:

T = 27^\circ\text{C} = 300

Use the osmotic pressure formula:

\pi = \frac{n}{V} R T = \frac{0.01}{0.1} \times 0.082 \times 300

Simplify stepwise:

\frac{0.01}{0.1} = 0.1

0.1 \times 0.082 \times 300 = 0.1 \times 24.6 = 2.46

atm

Answer:

The osmotic pressure of the solution is

2.46

atm, which corresponds to option A.

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