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Standard XII

Chemistry

Stoichiometric Calculations

20 mL of 0....

Question

$20$ mL of $0.2 M N a O H (a q)$ solution is mixed with $35$ mL of $0.1 M N a O H (a q)$ solution and the resultant solution is diluted to $100$ mL. $40$ mL of this diluted solution reacted with $10 %$ impure sample of oxalic acid $(H_{2} C_{2} O_{4})$ . The weight of impure sample is :

0.15

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0.135

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0.59

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none of the above

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Solution

The correct option is A $0.15$ g
$M_{N a O H}$ (resultant) $= \frac{20 \times 0.2 + 35 \times 0.1}{100} = 0.075 M$
Milli-equivalent of $N a O H =$ Milli-equivalent of $H_{2} C_{2} O_{4}$
Let weight of impure sample is $x$ g.
$40 \times 0.075 = \frac{x \times 0.90}{90} \times 2 \times 1000$
$x = 0.15 g$
Hence, the weight of impure sample is $0.15$ g.

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Similar questions

12.0

of an impure sample of arsenious oxide

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20 mL of 0.2MNaOH(aq) solution is mixed with 35 mL of 0.1MNaOH(aq) solution and the resultant solution is diluted to 100 mL. 40 mL of this diluted solution reacted with 10% impure sample of oxalic acid (H2C2O4). The weight of impure sample is :

The correct option is A 0.15 gMNaOH (resultant) =20×0.2+35×0.1100=0.075MMilli-equivalent of NaOH= Milli-equivalent of H2C2O4Let weight of impure sample is x g.40×0.075=x×0.9090×2×1000x=0.15gHence, the weight of impure sample is 0.15 g.

$20$ mL of $0.2 M N a O H (a q)$ solution is mixed with $35$ mL of $0.1 M N a O H (a q)$ solution and the resultant solution is diluted to $100$ mL. $40$ mL of this diluted solution reacted with $10 %$ impure sample of oxalic acid $(H_{2} C_{2} O_{4})$ . The weight of impure sample is :