Solutions Test

Result

Solutions Test - 9

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Weekly Quiz Competition

Question 1
1 / -0.25

The relative lowering in vapour pressure is proportional to the ratio of number of

A
Solute molecules to solvent molecules

B
Solvent molecules of solute molecules

C
Solute molecules to the total number of molecules in solution

D
Solvent molecules to the total number of molecules in solution

Solution
Question 2
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Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y?

A
20

B
50

C
100

D
200

Solution
Question 3
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An aqueous solution of 2% non - volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

A
23.4 g mol^-1

B
41.35 g mol^-1

C
10 g mol^-1

D
20.8 g mol^-1

Solution

Vapour pressure of pure water at boiling point (P^∘ ) = 1 atm = 1.013 bar
Vapour pressure of solution (P_s ) = 1.004 bar
Let Mass of solution = 100g
Mass of solute = (w) = 2g
Mass of solvent = 100 - 2 = 98g
Question 4
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In the graph plotted between vapour pressure (V.P) and temperature (T).

A
PQ is the curve for solvent, XY is the curve of solution and ΔT is depression on freezing point

B
PQ is the curve for solution, XY is the curve for solvent and ΔT is elevation in boiling point

C
PQ is the curve for solvent, XY is the curve for solution and ΔT is molal elevation in boiling point

D
PQ is the curve for solvent, XY is the curve for solution and ΔT is elevation in boiling point

Solution

PQ corresponds to increase in vapour pressure of solvent with temperature and XY corresponds to increase in vapour pressure of solution with temperature. ΔT is the elevation in boiling point of a solution.
Question 5
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A solution containing 12.5g of non-electrolyte substance in 185g of water shows boiling point elevation of 0.80K. Calculate the molar mass of the substance. (K_b = 0.52K kg mol⁻¹ )

A
53.06 g mol⁻¹

B
25.3 g mol⁻¹

C
16.08 g mol⁻¹

D
43.92 g mol⁻¹

Solution
Question 6
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If 1g of solute (molar mass = 50g mol⁻¹ ) is dissolved in 50g of solvent and the elevation in boiling point is 1K. The molar boiling constant of the solvent is?

A
2

B
3

C
2.5

D
5

Solution
Question 7
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2 g of sugar is added to one litre of water to give sugar solution. What is the effect of addition of sugar on the boiling point and freezing point of water?

A
Both boiling point and freezing point increase

B
Both boiling point and freezing point decrease

C
Boiling point increases and freezing point decreases

D
Boiling point decreases and freezing point increases

Solution

When a non-volatile solute is added to water there is elevation in boiling point and depression is freezing point.
Question 8
1 / -0.25

Sprinkling of salt helps in clearing the snow-covered roads in hills. The phenomenon involved in the process is

A
Lowering in vapour pressure of snow

B
Depression in freezing point of snow

C
Increase in freezing point of snow

D
Melting of ice due to increase in temperature by putting salt

Solution

When salt is spread over snow covered roads, snow starts melting from the surface because of the depression in freezing point of water and it helps in clearing the roads.
Question 9
1 / -0.25

Equimolar solutions in the same solvent have

A
Same elevation in boiling point and same depression in freezing point

B
Different elevation in boiling point and different depression in freezing point

C
Same elevation in boiling point but different depression in freezing point

D
Same depression in freezing point but different elevation in boiling point

Solution

Equimolal solutions show same colligative properties i.e., equal elevation in boiling point and equal depression in freezing point.
Question 10
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A 5% solution (w/W) of cane sugar (molar mass = 342 g mol^-1 ) has freezing point of 271 K. What will be the freezing point of a 5% glucose (molar mass = 18 g mol^-1 ) in water if freezing point of pure water is 273.15 K?

A
273.07 K

B
269.07 K

C
273.15 K

D
260.09 K

Solution

For cane sugar solution, 2.15 K =
For glucose solution,

Freezing point of glucose solution = 273.15 - 4.085 = 269.07 K