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Introduction to Oxidation and Reduction

20 g of a sam...

Question

$20 g$ of a sample of $B a (O H)_{2}$ is dissolved in $10 m L$ of $0.5 N H C l$ solution. The excess of $H C l$ was titrated with $0.2 N N a O H .$ The volume of $N a O H$ required was $10 m L$ . Calculate the percentage of $B a (O H)_{2}$ in the sample.
(Given molar mass of $B a$ is $137 g m o l^{- 1}$ )

12.8 %

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1.28 %

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50 %

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33.33 %

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Solution

The correct option is B $1.28 %$
$Milliequivalents of H C l present initially = 10 \times 0.5 = 5$
$Milliequivalents of NaOH consumed = Milliequivalents of H C l in excess = 10 \times 0.2 = 2$
$∴ Milliequivalents of H C l consumed = Milliequivalents of B a (O H)_{2} = 5 - 2 = 3$
$∴ Equivalents of B a (O H)_{2} = 3 / 1000 = 3 \times 10^{- 3}$
$Molecular Mass of B a (O H)_{2} = 137 + (16 + 1) \times 2 = 171 g m o l^{- 1}$
$Mass of B a (O H)_{2} = \frac{3 \times 10^{- 3} \times 171}{2} = 0.2565 g$
$% B a (O H)_{2} = \frac{0.2565}{20} \times 100 = 1.28 %$

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Similar questions

10 g

of a sample of

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is dissolved in

10 m L

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solution. The excess of HCl was titrated with

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20 c c

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20 g

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Introduction to Oxidation and Reduction

Standard XII Chemistry

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20 g of a sample of Ba(OH)2 is dissolved in 10 mL of 0.5 N HCl solution. The excess of HCl was titrated with 0.2 N NaOH. The volume of NaOH required was 10 mL. Calculate the percentage of Ba(OH)2 in the sample. (Given molar mass of Ba is 137 g mol−1)