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Ideal Gas Equation

1.25 g of a s...

Question

$1.25 g$ of a sample of ${N a}_{2} {C O}_{3}$ and ${N a}_{2} {S O}_{4}$ is dissolved in $250 m l$ solution. $25 m l$ of this solution neutralises $20 m l$ of $0.1 N$ $H_{2} {S O}_{4}$ . The % of ${N a}_{2} {C O}_{3}$ in this sample is:

84.8

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8.48

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15.2

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42.4

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Solution

The correct option is A $84.8$ %
The equivalent weight of ${N a}_{2} {C O}_{3}$ is 53.

Let the amount of ${N a}_{2} {C O}_{3}$ present in the mixture be $x$ $g$ . ${N a}_{2} {S O}_{4}$ will not react with $H_{2} {C O}_{3}$ .

Then, the number of gram equivalents of ${N a}_{2} {C O}_{3} =$ the number of gram equivalents of $H_{2} {S O}_{4}$ .

$\frac{x}{53} = \frac{20 \times 0.1 \times 10}{1000}$

$∴ x = 1.06 g$

$Note$ : We have divided with 1000 to convert the volume of $H_{2} {S O}_{4}$ from mL to L.
We have multiplied with 10 as out of 250 mL of ${N a}_{2} {S O}_{4}$ solution, only 25 mL is used for neutralisation with $H_{2} {S O}_{4}$ $\frac{250}{25} = 10$

The number of gram equivalents of $H_{2} {S O}_{4}$ are obtained by multiplying volume with normality.

The percentage of ${N a}_{2} {C O}_{3} = \frac{1.06 \times 100}{1.25} = 84.8$ %

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Similar questions

25 g m

mixture of

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and

{N a}_{2} {S O}_{4}

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using phenolphthalein indicator. The mass of

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1.20 g

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was exactly neutralised by

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l o g 5 = 0.6990

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250 m L

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2.65 g

{N a}_{2} {C O}_{3} \cdot 10 m L

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0.001 M

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[Molecular weight of

{N a}_{2} {C O}_{3} = 106

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1.25 g of a sample of Na2CO3 and Na2SO4 is dissolved in 250 ml solution. 25ml of this solution neutralises 20 ml of 0.1 N H2SO4. The % of Na2CO3 in this sample is:

$1.25 g$ of a sample of ${N a}_{2} {C O}_{3}$ and ${N a}_{2} {S O}_{4}$ is dissolved in $250 m l$ solution. $25 m l$ of this solution neutralises $20 m l$ of $0.1 N$ $H_{2} {S O}_{4}$ . The % of ${N a}_{2} {C O}_{3}$ in this sample is: