$(a)\;168.2ml\qquad(b)\;138.4ml\qquad(c)\;157.8ml\qquad(d)\;173.2ml$

Step 1:

Let a mole of $Na_2CO_3$ and a mole of $NaHCO_3$ are given :

Meq of $Na_2CO_3$+Meq of $NaHCO_3$=Meq of HCl

$a\times 2\times 1000+a\times 1\times 1000=0.1\times V$

$3a=10^{-4}\times V$-------(1)

Total mixture is 1g

Weight of $Na_2CO_3$+weight of $NaHCO_3=1g$

$a\times 106+a\times 84=1$

$a=5.26\times 10^{-3}$-----(2)

Step 2:

By equation(1) & (2)

$3\times 5.26\times 10^{-3}=10^{-4}\times V$

$V=157.8ml$

Hence (c) is the correct answer.

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