$(a)\;95\%\qquad(b)\;92\%\qquad(c)\;93\%\qquad(d)\;91\%$

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Volume of HCl used against $CaCO_3$ is not given.

Excess of HCl is neutralized by NaOH

$N_1V_1=N_2V_2$

$\Rightarrow \big(\large\frac{1}{10}\big)$$\times V_1=\big(\large\frac{1}{10}\big)$$\times 40$

$V_1=40ml$=Volume of excess HCl

Volume of HCl used for $CaCO_3=230-40=190cc$

Meq of HCl=Meq of $CaCO_3$

$\large\frac{1}{10}$$\times 190=\large\frac{g}{E}$$\times 1000$

g=0.95=95% Pure carbonate.

Hence (a) is the correct answer.

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