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pH of Salts

10 g of a mix...

Question

10 g of a mixture of barium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is precipitated according to the equation given below:

Na₂SO₄ + BaCl₂ → BaSO₄ +2NaCl

[O = 16, Na = 23, S = 32, Ba = 137]

Calculate the percentage of sodium sulphate in the original mixture.

A

42.6 %

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B

43.2 %

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C

44.8 %

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D

42.4 %

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E

46.2 %

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Solution

The correct option is A
42.6 %

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Q. 10 g of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g. of barium sulphate is precipitated according to the equation :-

N a_{2} S O_{4} + B a C l_{2} ⟶ B a S O_{4} + 2 N a C l

.
The percentage of sodium sulphate in the original mixture is:

[N a = 23, O = 16, S = 32, B a = 132]

Q. 1.42 g of sodium sulphate is dissolved in 100 mL water. 4.16 g of barium chloride is dissolved in another 100 mL water. The amount of precipitate of barium sulphate upon mixing of the two solutions is:

N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 N a C l

(Atomic mass: Na = 23, S = 32, O =16, Ba = 137, Cl = 35.5)

Q. 1.42 g of sodium sulphate is dissolved in 100 mL water. 4.16 g of barium chloride is dissolved in another 100 mL water. The amount of precipitate of barium sulphate upon mixing of the two solutions is:

N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 N a C l

(Atomic mass: Na = 23, S = 32, O =16, Ba = 137, Cl = 35.5)

0.98

g of the metal sulphate was dissolved in water and excess of barium chloride was added. The precipitated barium sulphate weighted

0.95

g. Calculate the equivalent weight of the metal.

Q. 1.42 g of sodium sulphate dissolved in water was treated with excess of

B a C l_{2}

solution and 1.167 g of

B a S O_{4}

was obtained. Calculate the percentage yield of the reaction.
Molar mass of sodium sulphate and barium sulphate is 142 g/mol and 233.4 g/mole respectively.

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