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Anomalous Solubility

20 ml of 0.1 ...

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20 ml of 0.1 M Bacl2 reacts with 30 ml of 0.2 M Al2(SO4)3 then what is weight of BaSO4 formed ?

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Q. Calculate the amount of precipitate of

B a S O_{4}

formed when 30 mL of 0.1 M

B a C l_{2}

is mixed with 40 mL of 0.2 M

A l_{2} (S O_{4})_{3}

.
(Molar mass of barium = 137 g/mol, Molar mass of Al = 27 g/mol).

Q. 50 mL solution of

{BaCl}_{2}

(20.8% w/v) and 100 mL solution of

H_{2} {SO}_{4}

(9.8% w/v) are mixed then maximum mass of

{BaSO}_{4}

[{BaCl}_{2} {+ H}_{2} {SO}_{4} \to {BaSO}_{4} + 2HCl]

Q. 25 mL of 0.15 M Pb

(N O_{3})_{2}

reacts completely with 20 mL of

A l_{2} (S O_{4})_{3}

. The molar concentration of

A l_{2} (S O_{4})_{3}

3 P b (N O_{3})_{2} (a q) + A l_{2} (S O_{4})_{3} (a q .) \to 3 P b S O_{4} (s) + 2 A l (N O_{3})_{3} (a q .)

20

1 M B a C l_{2}

25

0.1

N a_{2} S O_{4}

are mixed to form

B a S O_{4}

. The amount of

B a S O_{4}

precipitated are:

Q. 100 mL of 20.8%

B a C l_{2}

solution and 50 mL of 9.8%

H_{2} S O_{4}

solution will form

B a S O_{4}

(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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