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.

221. A 0.1539 molal aqueous solution of cane sugar (mol. mass = 342 g mol^–1) has a freezing point of 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of gluocose (mol. mass = 180 g mol^–1) per 100 g of solution?

187 Views

222.

An aqueous solution of 2 percent non-volatile solute exerts a pressure of 0.096 atm at the boiling poine of the solvent. What is the molecular mass of the solute ?

155 Views

223. The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of a one molal solution of a non-volatile, non-ionic solute in water.

215 Views

224. Why is the vapour pressure of a solution of glucose in water lower than that of water?

477 Views

225. A 6.90 M solution of KOH in water contains 30% by mass of KOH. Calculate the density of the KOH solution. [Molar mass of KOH = 56 g mol^–1]

214 Views

226. Give reasons for the following:
When 30 ml of ethyl alcohol and 30 ml of water are mixed, the volume of resulting solution is more than 60 ml.

188 Views

227. Give reasons for the following:
Copper is conducting as such while copper sulphate is conducting only in molten state or in aqueous solution.

822 Views

228. Urea forms an ideal solution in water. Determine the vapour pressure of an aqueous solution containing 10% by mass of urea at 40⁰C. (Vapour pressure of water at 40⁰C = 55.3 mm of Hg)

Answer:

by using formula .

$\frac{p ° - p}{p °} = x_{u r e a}$

let the mass of solution be 100g
therefore mass of urea = 10g
molecular mass of urea = 60 g

x_urea = 10/60 = 1/6

molecular mass of water = 18 g
x_water = 90/18 = 5

mole fraction of urea = $x_{urea} = \frac{1 / 6}{1 / 6 + 5} = \frac{1 / 6}{31 / 6} x_{urea} = 1 / 31 using above formula \frac{55.3 - p}{55.3} = \frac{1}{31} 55.3 - p = \frac{55.3}{31} 55.3 - \frac{55.3}{31} = p 55.3 (1 - \frac{1}{31}) = p p = 55.3 \times \frac{30}{31} = 53.52 mm Hg$

559 Views

229. Why is freezing point depression of 0.1 M sodium chloride solution nearly twice that of 0.1 M glucose solution?

2697 Views

230. A 6.90 M solution of KOH in water contains 30% by mass of KOH. Calculate the density of the KOH solution. (Molar mass of KOH = 56 g mol^–1)

204 Views

Previous Year Papers

Test Series

Test Yourself

Short Answer Type

228. Urea forms an ideal solution in water. Determine the vapour pressure of an aqueous solution containing 10% by mass of urea at 400C. (Vapour pressure of water at 400C = 55.3 mm of Hg)

228. Urea forms an ideal solution in water. Determine the vapour pressure of an aqueous solution containing 10% by mass of urea at 40⁰C. (Vapour pressure of water at 40⁰C = 55.3 mm of Hg)