0.7 g of Na_2CO_3 cdot xH_2O were dissolved | vedclass

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(C) The reaction between $Na_2CO_3$ and $HCl$ is: $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$. The number of equivalents of $Na_2CO_3$ in $20 \

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$2$

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(C) The reaction between $Na_2CO_3$ and $HCl$ is: $Na_2CO_3 + 2HCl ightarrow 2NaCl + H_2O + CO_2$. The number of equivalents of $Na_2CO_3$ in $20 \ mL$ is equal to the number of equivalents of $HCl$ used. Equivalents of $HCl = N 	imes V(L) = \frac{1}{10} 	imes \frac{19.8}{1000} = 1.98 	imes 10^{-3} \ eq$. Since $20 \ mL$ contains $1.98 	imes 10^{-3} \ eq$,the total $100 \ mL$ contains $1.98 	imes 10^{-3} 	imes \frac{100}{20} = 9.9 	imes 10^{-3} \ eq$. The equivalent weight of $Na_2CO_3$ is $\frac{106}{2} = 53 \ g/eq$. Total mass of $Na_2CO_3$ (anhydrous) = $9.9 	imes 10^{-3} 	imes 53 = 0.5247 \ g$. The mass of $Na_2CO_3 \cdot xH_2O$ is $0.7 \ g$. The fraction of $Na_2CO_3$ in the hydrate is $\frac{106}{106 + 18x} = \frac{0.5247}{0.7} \approx 0.7495$. $106 = 0.7495 	imes (106 + 18x) \implies 106 = 79.45 + 13.49x \implies 13.49x = 26.55 \implies x \approx 2$.

$0.7 \ g$ of $Na_2CO_3 \cdot xH_2O$ were dissolved in water and the volume was made to $100 \ mL$. $20 \ mL$ of this solution required $19.8 \ mL$ of $N/10 \ HCl$ for complete neutralization. The value of $x$ is:

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