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Standard XII

Chemistry

Stoichiometric Calculations

One gm of i...

Question

One $g m$ of impure sodium carbonate is dissolved in water and the solution is made up to $250 m l$ . To $50 m l$ of this made up solution, $50 m l$ of $0.1 N$ $H C l$ is added and the mixture, after shaking well required $10 m l$ of $0.16 N - N a O H$ solution for complete titration.

Calculate the $%$ purity of the sample:

90.1 %

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20.1 %

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92.1 %

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90.2 %

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Solution

The correct option is A $90.1 %$
M eq. of $N a_{2} C O_{3} =$ M eq. of $H C l$
$\frac{w}{\frac{106}{2}} \times 1000 = 50 \times 0.1 - 10 \times 0.16$
$%$ purity $= \frac{w}{1} \times 100 = 90.1 %$

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One gm of impure sodium carbonate is dissolved in water and the solution is made up to 250 ml. To 50 ml of this made up solution, 50 ml of 0.1 N HCl is added and the mixture, after shaking well required 10 ml of 0.16 N−NaOH solution for complete titration. Calculate the % purity of the sample:

The correct option is A 90.1%M eq. of Na2CO3= M eq. of HCl w1062×1000=50×0.1−10×0.16 % purity =w1×100=90.1%

The correct option is A $90.1 %$
M eq. of $N a_{2} C O_{3} =$ M eq. of $H C l$
$\frac{w}{\frac{106}{2}} \times 1000 = 50 \times 0.1 - 10 \times 0.16$
$%$ purity $= \frac{w}{1} \times 100 = 90.1 %$