The density of 3M solution of sodium thiosulphate (Na 2 S 2 O 3 ) is 1.58 g/ml. Calculate (i) amount of Na 2 S 2 O 3 in % w/w (ii) mole fraction of Na 2 S 2 O 3 (iii) molality of Na + and S 2 O 3 2– ions.

Given : 3 M solution of Sodium thiosulphate ( 3 × 158 g / l ) density of solution = 1 . 58 g / mL volume of solution = 1 L mass of solution = volume × density = 1000 × 1 . 58 = 1580 g m mass of solvent = mass of solution - mass of solute = 1580 - 3 × 158 = 1106 g ( I ) Mass % of solution = mass of solute mass of solution × 100 = 3 × 158 1580 × 100 = 30 % (ii) since mass % of solution is 30 % , hence mass % of solvent is 70 % mole fraction of Na 2 S 2 O 3 = moles of Na 2 S 2 O 3 total moles = 30 / 158 70 / 18 + 30 / 158 = 0 . 046 (iii) Molality of solution = n o . o f m o l e s o f s o l u t e m a s s o f s o l v e n t ( K g ) = 3 1580 - 3 × 158 × 1000 = 2 . 71 m therefore molality of Na + = 2 × 2 . 71 and molality of S 2 O 3 2 - = 2 . 71 (i) 30 % , (ii) 0 . 046 (iii) Na + = 5 . 42 m , S 2 O 3 2 – = 2 . 71 m

The density of 3M solution of sodium thiosulphate (Na 2 S 2 O 3 ) is 1.58 g/ml. Calculate (i) amount of Na 2 S 2 O 3 in % w/w (ii) mole fraction of Na 2 S 2 O 3 (iii) molality of Na + and S 2 O 3 2– ions.

Given : 3 M solution of Sodium thiosulphate ( 3 × 158 g / l ) density of solution = 1 . 58 g / mL volume of solution = 1 L mass of solution = volume × density = 1000 × 1 . 58 = 1580 g m mass of solvent = mass of solution - mass of solute = 1580 - 3 × 158 = 1106 g ( I ) Mass % of solution = mass of solute mass of solution × 100 = 3 × 158 1580 × 100 = 30 % (ii) since mass % of solution is 30 % , hence mass % of solvent is 70 % mole fraction of Na 2 S 2 O 3 = moles of Na 2 S 2 O 3 total moles = 30 / 158 70 / 18 + 30 / 158 = 0 . 046 (iii) Molality of solution = n o . o f m o l e s o f s o l u t e m a s s o f s o l v e n t ( K g ) = 3 1580 - 3 × 158 × 1000 = 2 . 71 m therefore molality of Na + = 2 × 2 . 71 and molality of S 2 O 3 2 - = 2 . 71 (i) 30 % , (ii) 0 . 046 (iii) Na + = 5 . 42 m , S 2 O 3 2 – = 2 . 71 m

Calculate the molality and molarity of a solution made by mixing equal volumes of 30 % by weight of H 2 SO 4 (density = 1 . 20 g / mL ) and 70 % by weight of H 2 SO 4 (density = 1 . 60 g / mL ).

Given, Two Solutions: Sol A = 30 % (w/W) of density = 1 . 20 g / mL of H 2 SO 4 . Sol B = 70 % (w/W) of density = 1 . 60 g / mL of H 2 SO 4 . Molality of solution is the number of moles of solute mass of solvent ( Kg ) Molarity of solution is the number of moles of solute volume of solution ( L ) Mass of solvent = Mass of solution − Mass of solute. For A = ( volume × density ) of solution − 30 % by weight of solution = 100 × 1 . 20 − 0 . 3 × 120 = 84 g . For B = ( volume × density ) of solution − 70 % by weight of solution = 100 × 1 . 60 − 0 . 7 × 160 = 48 g . Molality of solution A + B = no . of moles of solute A mass of solvent ( Kg ) + no . of moles of solute B mass of solvent ( Kg ) = 36 / 98 + 112 / 98 84 + 48 × 1000 = 11 . 44 m . Let, the volume of each solution is 100 mL . Molarity of A + B = no . of moles of solute A volume of solution ( L ) + no . of moles of solute B volume of solution ( L ) = 39 / 98 + 112 / 98 200 × 1000 = 7 . 55 M . Molality = 11 . 44 m , Molarity = 7 . 55 M

Two liquids, A & B , form an ideal solution. At the specified temperature, the vapour pressure of pure A is 200 mm Hg while that of pure B is 75 mm Hg . If the vapour over the mixture consists of 50 mol % A , what is the mole per cent A in the liquid?

Given, in the vapour phase mole fraction of liquid A = 50 % = 50 100 = 0 . 5 Therefore, the mole fraction of liquid B = 1 - 0 . 5 = 0 . 5 Y A = P A X A P t Y B = P B X B P t Y A = Y B P A = 200 , P B = 75 P A X A P B = X B 200 X A 75 = 1 - X A 8 X A 3 = 1 - X A on solving X A = 3 11 = 0 . 27 in percent 0 . 27 × 100 = 27 %

Choose the correct statement with respect to the vapour pressure of a liquid among the following :

Increases non-linearly with increasing temperature

Decreases non-linearly with increasing temperature

Decreases linearly with increasing temperature

Increases linearly with increasing temperature

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The density of 3M solution of sodium thiosulphate (Na₂S₂O₃) is 1.58 g/ml. Calculate

(i) amount of Na₂S₂O₃in % w/w

(ii) mole fraction of Na₂S₂O₃

(iii) molality of Na⁺ and S₂O₃ ^2– ions.

Important Questions on Solutions

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Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 10

Calculate the molality and molarity of a solution made by mixing equal volumes of

30 %

by weight of

(I) Mass % of solution

(density

= 1.20 g / mL

) and

70 %

by weight of

H_{2} {SO}_{4}

(density

= 1.60 g / mL

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Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 13

A mixture of an organic liquid

A

and water is distilled under

since mass % of solution is 30 %, hence mass % of solvent is 70 %

99.2 ° C

. How many grams of steam will be condensed to obtain

1.0 g

of liquid

A

in the distillate? (Vapour pressure of water at

99.2 ° C

739 mm Hg

and molecular weight of

A = 123

)

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Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 14

The boiling point of a mixture of water and nitrobenzene is

99 ° C

, the vapour pressure of water is

733 mm of Hg

and the atmospheric pressure is

760 mm of Hg

. The molecular weight of nitrobenzene is

123

. Find the ratio of weights of the components of the distillate.

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Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 15

Two liquids

A & B

form an ideal solution. At

300 K

, the vapour pressure of a solution containing

1 mole of A

and

3 moles of B

550 mm of Hg

. At the same temperature, if one mole of

B

is added to this solution, the vapour pressure of the solution increases by

10 mm of Hg

. Determine the vapour pressure of

A and B

in their pure states.

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Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 16

Two liquids,

A & B

, form an ideal solution. At the specified temperature, the vapour pressure of pure

A

=

while that of pure

B

75 mm Hg

. If the vapour over the mixture consists of

50 mol % A

, what is the mole per cent

A

in the liquid?

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Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 17

Two solutions of

A & B

are available. The first is known to contain

1 mole of A & 3 moles of B

and its total vapour pressure is

1.0 atm

. The second is known to contain

2 moles of A & 2 moles of B

its vapour pressure is greater than

1 atm

, but it is found that this total vapour pressure may be reduced to

1 atm

by the addition of

6 moles of C

. The vapour pressure of pure

C

0.80 atm

. Assuming ideal solutions and that all these data refer to

25^{0} C

, calculate the vapour pressure of pure

A

and of pure

B

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Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 18

80^{0} C

, the vapour pressure of pure benzene is

753 mm Hg

and of pure toluene

290 mm Hg

. Calculate the composition of a liquid in mole per cent at

80^{0} C

in equilibrium with the vapour containing.

30 mole %

of benzene.

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Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 19

The vapour pressure of

C_{6} H_{6}

and

C_{7} H_{8}

mixture at

50^{0} C

is given by

P (mm Hg) = 180 X_{B} + 90

, where

X_{B}

is the mole fraction of

C_{6} H_{6}

. A solution is prepared by mixing

12 mol C_{6} H_{6}

and

8 mol C_{7} H_{8}

and if vapours over this solution are removed and condensed into liquid and again brought to the temperature

50^{0} C

, what would be the mole fraction of

C_{6} H_{6}

in the vapour state? (Atomic weight of

C = 12, H = 1

)

The density of 3M solution of sodium thiosulphate (Na2S2O3) is 1.58 g/ml. Calculate (i) amount of Na2S2O3 in % w/w (ii) mole fraction of Na2S2O3 (iii) molality of Na+ and S2O3 2– ions.

Important Questions on Solutions

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The density of 3M solution of sodium thiosulphate (Na₂S₂O₃) is 1.58 g/ml. Calculate

(i) amount of Na₂S₂O₃in % w/w

(ii) mole fraction of Na₂S₂O₃

(iii) molality of Na⁺ and S₂O₃ ^2– ions.