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How many ml of 0.1N HCl are required to react completely with 1g mixture of $Na_2CO_3$ and $NaHCO_3$ containing equimolar amount of two ?


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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)
$3\times 5.26\times 10^{-3}=10^{-4}\times V$
Hence (c) is the correct answer.
answered Nov 6, 2013 by sreemathi.v

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