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Chapter 2 Solutions Solutions

Question - 41 : -

Calculate the amount of benzoicacid (C6H5COOH) required for preparing 250 mL of 0.15 Msolution in methanol.

Answer - 41 : -

0.15 M solution of benzoic acidin methanol means,

1000 mL of solution contains 0.15mol of benzoic acid

Therefore,250 mL of solution contains =  mol of benzoic acid

= 0.0375 mol of benzoic acid

Molar mass of benzoic acid (C6H5COOH)= 7 × 12 + 6 × 1 + 2 × 16

= 122 g mol−1

Hence, required benzoic acid =0.0375 mol × 122 g mol−1

=4.575 g

Question - 42 : -

The depression in freezing pointof water observed for the same amount of acetic acid, trichloroacetic acid andtrifluoroacetic acid increases in the order given above. Explain briefly.

Answer - 42 : -

Among H, Cl, and F, H is leastelectronegative while F is most electronegative. Then, F can withdraw electronstowards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H+ ionsi.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ionsproduced, the greater is the depression of the freezing point. Hence, thedepression in the freezing point increases in the order:

Acetic acid < trichloroaceticacid < trifluoroacetic acid

Question - 43 : -

Calculate the depression in thefreezing point of water when 10 g of CH3CH2CHClCOOHis added to 250 g of water. Ka = 1.4 × 10−3K=1.86K kg mol−1.

Answer - 43 : -

Molalityof the solution, 
Let α be the degree of dissociationof 
undergoes dissociation accordingto the following equation:

Question - 44 : -

19.5 g of CH2FCOOH isdissolved in 500 g of water. The depression in the freezing point of waterobserved is 1.0°C. Calculate the van’t Hoff factor and dissociation constant offluoroacetic acid.

Answer - 44 : -

Question - 45 : -

Vapour pressure of water at 293Kis 17.535 mm Hg. Calculate the vapour pressure of water at 293 Kwhen 25 g ofglucose is dissolved in 450 g of water.

Answer - 45 : -

Question - 46 : -

Henry’s law constant for themolality of methane in benzene at 298 Kis 4.27 × 105 mm Hg.Calculate the solubility of methane in benzene at 298 Kunder 760 mm Hg.

Answer - 46 : -

Question - 47 : -

100 g of liquid A (molar mass 140g mol−1) was dissolved in 1000 g of liquid B (molar mass 180 g mol−1). 

Answer - 47 : - The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.


Answer

Question - 48 : -

Benzene and toluene form idealsolution over the entire range of composition. 

Answer - 48 : - The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Answer

Question - 49 : - The air is a mixture of a number of gases. The major components are oxygen

Answer - 49 : -

and nitrogen with approximateproportion of 20% is to 79% by volume at 298

K. The water is in equilibriumwith air at a pressure of 10 atm. At 298 Kif the

Henry’s law constants for oxygenand nitrogen are 3.30 × 107 mm and 6.51 × 107 mmrespectively, calculate the composition of these gases in water.

Answer

Question - 50 : -

Determine the amount of CaCl2 (=2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75atm at 27°C.

Answer - 50 : -

Hence, the required amount of CaCl_2 is 3.42 g.


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