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

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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.

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