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0.5g of a mixture of $K_2CO_3$ and $Li_2CO_3$ required 30ml of 0.25N HCl solution for neutralization.What is % composition mixture ?

$(a)\;80\%,20\%\qquad(b)\;90\%,10\%\qquad(c)\;91\%,9\%\qquad(d)\;96\%,4\%$

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Step 1:
Weight of $K_2CO_3=ag$
Weight of $Li_2CO_3=bg$
$a+b=0.5$------(1)
For reaction
Meq.of $K_2CO_3$+Meq of $Li_2CO_3$=Meq of HCl
$\large\frac{a\times 1000}{138/2}+\frac{b\times 1000}{74/2}$$=30\times 0.25$
$74a+138b=38.295$------(2)
Step 2:
By equation (1) & (2)
$a=0.48g$
$b=0.02g$
$\%$ of $K_2CO_3=\large\frac{0.48}{0.5}$$\times 100$
$\Rightarrow 96\%$
$\%$ of $Li_2CO_3=\large\frac{0.02}{0.5}$$\times 100$
$\Rightarrow 4\%$
Hence (d) is the correct answer.
answered Nov 6, 2013 by sreemathi.v
 

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