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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 %
$N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} ↓ + 2 N a C l$
$Moles of B a S O_{4} = \frac{given mass}{molar mass} = \frac{11.65}{233} = 0.05 mol$
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 = $0.05$ mol $\times 142$ $(\frac{g}{mol}) = 7.1 g$
$Percentage purity = \frac{mass of pure compound in sample}{total mass of impure sample} \times 100 = \frac{7.1}{10} \times 100$
$= 71 %$

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Similar questions

Q. 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}

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N a_{2} S O_{4}

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

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Percentage Composition

Standard XI 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)