The amount of mL of $0.1 N H C l$ required to react completely with $1$ g mixture of ${N a}_{2} C O_{3}$ and $N a H C O_{3}$ containing equimolar amounts of the two is (as the nearest integer) :

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Solution

Let the moles of ${N a}_{2} C O_{3}$ and $N a H C O_{3}$ be $a$ each.
Meq. of ${N a}_{2} C O_{3}$ + Meq. of $N a H C O_{3} =$ Meq. of $H C l$
$a \times 2 \times 1000 + a \times 1 \times 1000 = 0.1 \times V$
$∴ 3 \times a = 10^{- 4} \times V . . . (i)$
Since, mass of mixture $= 1$ g
$∴$ Mass of ${N a}_{2} C O_{3}$ + Mass of $N a H C O_{3} = 1$ g
or $a \times 106 + a \times 84 = 1$
$∴ a = 5.26 \times 10^{- 3} . . . (i i)$
From Eqs. $(i)$ and $(i i)$ , we get $3 \times 5.26 \times 10^{- 3} = 10^{- 4} \times V$
or $V = 157.8$ mL
So, the answer is $158$ mL.

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