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5.3 grams of ${Na}_{2} {CO}_{3}$ is dissolved in 200ml of a solution. It is diluted by adding 800 ml of water. 100ml of the resulting solution requires ‘V’ ml of 0.02N $H_{2} {SO}_{4}$ solution for complete neutralization. ‘V’ is

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250 ml

100ml

500ml

2000ml

answer is C.

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Detailed Solution

i)    $N = \frac{5 .3}{53} \times \frac{1000}{200}$
N = 0.5N
ii) $V_{1} N_{1} = V_{2} N_{2}$
$200 (0 .5) = (200 + 800) N_{2}$
   $N_{2} = 0 .1 N$
iii) $V_{b} N_{b} = V_{a} N_{a}$
   $100 (0 .1) = V_{a} (0 .02)$
$V_{a} = 500 ml$

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