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Standard XII

Chemistry

Percentage Composition

11.65 g of Ba...

Question

11.65 g of $B a S O_{4}$ was obtained as a dry precipitate when 10 g of an impure sample of $N a_{2} S O_{4}$ was dissolved in water and treated with an excess of $B a C l_{2}$ . Calculate the percentage purity of the $N a_{2} S O_{4}$ sample. (Molar mass of $B a S O_{4} = 233 g a n d N a_{2} S O 4 = 142 g$ )

38 %

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71 %

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22 %

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88 %

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Solution

The correct option is B 71 %
$N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} ↓ + 2 N a C l$
Moles of $B a S O_{4} = \frac{given mass}{molar mass} = \frac{11.65}{233} = 0.05$ moles
According to the reaction, 1 mole of $B a S O_{4}$ precipitate is obtained from 1 mole of $N a_{2} S O_{4}$
0.05 moles of $B a S O_{4}$ will be produced by 0.05 mole of $N a_{2} S O_{4}$
Amount of pure $N a_{2} S O_{4}$ required = 0.05 moles $\times$ 142 $(\frac{g}{mole}) = 7.1$ g
% purity = $\frac{mass of pure compound in sample}{total mass of impure sample} \times 100 = \frac{7.1}{10} \times 100$
= 71 %

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Similar questions

11.65 g

B a S O_{4}

was obtained as a dry precipitate when

10 g

of an impure sample of

N a_{2} S O_{4}

was dissolved in water and treated with an excess of

B a C l_{2}

. Calculate the percentage purity of the

N a_{2} S O_{4}

sample. (Molar mass of

B a S O_{4} = 233 g a n d N a_{2} S O 4 = 142 g

)

Q. 1000 g of a mixture of

N a_{2} C O_{3}, N a_{2} S O_{4}

and

N a O H

for complete neutralisation requires 511 g of

H C l

. The same mixture when reacted with excess of

B a C l_{2}

solution, produces 466 g of white precipitate of

B a S O_{4}

. Calculate the mass percentage of

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(Given Molar mass of

N a_{2} C O_{3} = 106 g, N a_{2} S O_{4} = 142 g, B a S O_{4} = 233 g, H C l = 36.5 g a n d N a O H = 40 g

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

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Q. The impure

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Percentage Composition

Standard XII Chemistry

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11.65 g of BaSO4 was obtained as a dry precipitate when 10 g of an impure sample of Na2SO4 was dissolved in water and treated with an excess of BaCl2. Calculate the percentage purity of the Na2SO4 sample. (Molar mass of BaSO4=233 g and Na2SO4=142 g)