To 10 mL of 1 M $B a {C l}_{2}$ solution 5 mL of 0.5 M $K_{2} S O_{4}$ is added. $B a S O_{4}$ is precipitated out.What will happen?

Freezing point will increase.

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Boiling point will increase.

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Freezing point will lower down.

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Boiling point will lower down.

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Solution

The correct options are
A Boiling point will increase.
C Freezing point will lower down.
Adding $K_{2} S O_{4}$ into $B a {C l}_{2}$ solution will result in precipitation of $B a S O_{4}$ which is non-volatile, and therefore boiling point of solution will be increased and its freezing point will decrease.
Hence, options B and C are correct.

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10

1 M

B a C l_{2}

5

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B a S O_{4}

10 ml of 1M Bacl2 solution and 5ml 0.5 M K2SO4 are mixed together to precipitate out BaS04 .The amount of BaSO4 precipited will be

a) 0.005.mol b) 0.00025mol

c)0.025mol. c)0.0025 mol

K_{2} S O_{4}

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(B a = 137, C l = 35.5, S = 32, H = 1, O = 16)

B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l

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