IIT JEE Questions And Answers :: Solutions: part2



IIT JEE : Solutions QUESTIONS AND ANSWERS :: part2 : 16 to 20

Following IIT JEE Entrance Multiple choice objective type questions and answers will help you in Maths, Physics and Chemistry 2024 examinations :

16.On dissolving 8.8 gm of non-volatile solute to 100 gm of water the vapour pressure of ice becomes 4.559 mm Hg. If vapour pressure of pure ice is 4.58 mm Hg than molecular weight of the substance should be

345.46
245.46 gm
34.576 gm
360

17.At constant temperature, the concentration of a slightly soluble gas in a liquid is directly proportional to the partial pressure of the gas. This is in accordance with

Joules law
Vant Hoff law
Henrys law
Mass Action

18.Molality of a solution is the number of moles of solute in

1 kg of solvent
1 litre of solvent
1 litre of solute
1 mole of solute

19.When 18 g of a compound was dissolved in 100 g of water, the vapour pressure of the latter was lowered from 17.54 mm to 17.23 mm of mercury. The molecular mass of compound is

2900
368
184
92

20.The vapour pressure of water at room temperature is 30 mm. If the mole fraction of the solvent in solution is 0.9, the vapour pressure of aqueous solution at room temperature will be

18 mm
36 mm
72 mm
27 mm

More Solutions QUESTIONS AND ANSWERS available in next pages

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