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(C) The reaction between $CaCO_3$ and $HCl$ is: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$. At the equivalence point,the number of milliequivale

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(C) The reaction between $CaCO_3$ and $HCl$ is: $CaCO_3 + 2HCl ightarrow CaCl_2 + H_2O + CO_2$. At the equivalence point,the number of milliequivalents $(M.eq.)$ of $HCl$ is equal to the number of milliequivalents of $CaCO_3$. $M.eq. 	ext{ of } HCl = M.eq. 	ext{ of } CaCO_3$. $N 	imes V(mL) = \frac{	ext{mass}}{	ext{equivalent mass}} 	imes 1000$. The equivalent mass of $CaCO_3$ is $\frac{100}{2} = 50 \ g/eq$. $N 	imes 50 = \frac{1.0}{50} 	imes 1000$. $N = \frac{1000}{50 	imes 50} = \frac{1000}{2500} = 0.4 \ N$.

$1.0 \ g$ of pure calcium carbonate was found to require $50 \ mL$ of dilute $HCl$ for complete reaction. The strength of the $HCl$ solution is given by ........ $N$.

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