$3.65 gm$ equimolar mixture of $NaOH$ and ${Na}_{2} {CO}_{3}$ is titrated against $0.1 M$ $HCI$ using phenolphthalein as an indicator, $V_{1} mL$ of acid was required to reach end point. In another experiment $3.65 gm$ of same mixture is titrated against $0.2 M HCl$ using methyl orange as an indicator. $V_{2} mL$ of acid was required to reach end point. $V_{1} + V_{2}$ is:

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875 mL

760 mL

500 mL

1000 mL

answer is A.

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Detailed Solution

$NaOH$ ${Na}_{2} {CO}_{3}$

a molar a molar

As per the first titration,

$2 a = 0.1 \times V_{1}$ . $∴ V_{1} = 0.50$ .............. (i)

In the second experiment,

$a + 2 a = 0.2 \times V_{2}$

$3.65 = a \times 40 + a \times 106$

$a = 0.025$

Putting the value of a in (1) we get,

$And V_{2} = 0.375$

$V_{1} + V_{2} = 0.50 + 0.375 = 0.875 ml$

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