Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Solutions

Short Answer Type

1. Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

1703 Views

Long Answer Type

Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

897 Views

3. Calculate the molarity of each of the following solution (a) 30 g of Co(NO₃)₂.6H₂O in 4.3 L solution (b) 30 mL of 0.5 MH₂SO₄ diluted to 500 mL.

844 Views

4. Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg 0.25 of molal aqueous solution.

1475 Views

Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.

1010 Views

Short Answer Type

6. H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at S.T.P. is 0.195 m, calculate Henry's law constant.

410 Views

7. Henry’s law constant for CO₂ in water is 1.67 x 10⁸ P_a at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.

479 Views

8. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg at 350 K respectively. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

223 Views

Long Answer Type

9. Vapour pressure of pure water at 298 K is 23.8 mm. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Vapoure pressure of pure water (solvent) at 298 K, p⁰ = 23.8 mm
Vapour pressure of solution, p = ?
Mass of solvent ,W = 850 g
Mass of solute,M = 50 g
Mol. mass of water (H₂O), M = 18 g mol^–1Mol.mass of urea NH₂ CO NH₂= 14 + 2 + 12 + 16 + 14 + 2
= 60 g mol^–1According to Raoult's law, $\frac{p^{0} - p}{p^{0}} = \frac{ωM}{Wm}$

$p = p^{0} - \frac{w \times M}{m \times W} \times p °$

$p = 23.8 - \frac{50 \times 18}{60 \times 850}$

$= 23.8 - 0.017 = 23.78$
Hence, 23.78 mm Hg. Ans.