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 Multiple Choice QuestionsShort Answer Type

231. Urea forms an ideal solution in water. Determine the vapour pressure of an aqueous solution contains 10% by mass of urea at 400C. (Vapour pressure of water at 400C = 55.3 mm of Hg)
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232. A solution is made by dissolving 30 g of a non-volatile solute in 90g of water. It has a vapour pressure of 2.8 kPa at 298 K. At 298 K, vapour pressure of pure water is 3.64 kPa. Calculate the molar mass of the solute.
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233. At 300 K, 36 g of glucose present per litre in its solution has an osmotic pressure of 4.98 bar. If osmotic pressure of solution is 1.52 bar at the same temperature, what would be its concentration.
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234.

Calculate the vapour- pressure lowering of water when5.67g of glucose, C6H12O6 is dissolved in 25.2 g water at 250C. The vapour pressure of water at 250C is 23.8 mmHg. What is the vapour pressure of the solution?

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235. A 4% solution of sucrose is isotonic with 3% solution of an unknown organic substance. Calculate the molecular mass of unknown resistance.
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236. Calculate molarity and molality of 13% solution (by weight of sulphuric acid. Its density is 1.020 g cm–3. (At. mass, H = 1,16, S = 32 amu).
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237.

What will be the boiling point of bromine when 87.25 x 10-3 g of Sulphur is added to 39g of bromine? Bromine boils at 332.15K [Kbfor Br2 -5.2K kgmole-1]

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238.

The freezing point of a solution containing 0.2g of acetic acid in 20.0 g of benzene is lowered by 0.450C. calculate the degree of association of acetic acid in benzene. (Kf for benzene =5.12K kg mol-1)

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239.

Calculate the amount of ice that will separate out on cooling a solution containing 50g of ethylene glycol in 200g water -9.30C (Kf for water =1.86K kf mol-1).

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 Multiple Choice QuestionsLong Answer Type

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240. What is osmotic pressure? How would you determine the molecular mass of solute with the help of osmotic pressure?


Answer:
osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus:
Π = C R T 
Here Π is the osmotic pressure and R is the
gas constant.
Π = (n2 /V) R T
Here V is volume of a solution in litres containing n2 moles of solute.
If w2 grams of solute, of molar mass, M2 is present in the solution, then
n2 = w2 / M2 and we can write,

πV= w2RTM2

Thus, knowing the quantities w2, T, Π and V we can calculate the molar mass of the solute.

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