Question

$0.7$ grams of $N{a}_{2}C{O}_3\cdot x{H}_{2}O$ were dissolved in water and the volume was made to $100mL,20mL$ of this solution required $19.8mL$ of $N/10HCl$ for complete neutralization. The value of $x$ is Report your answer by rounding it upto nearest whole number.

Answer 1

Answer: 2

Meq. of $N{a}_{2}C{O}_3\cdot x{H}_{2}O$ in $20mL=19.8\times \frac{1}{10}$
$\therefore$ Meq. of $N{a}_{2}C{O}_3\cdot x{H}_{2}ON{a}_{2}C{O}_3\cdot x{H}_{2}O$ in $100mL=19.8\times \frac{1}{10}\times 5$
$\therefore \frac{w}{E}\times 1000=19.8\times \frac{1}{10}\times 5$
or $\frac{0.7}{\frac{M}{2}}\times 1000=\frac{19.8}{2}$
$\therefore M=141.41$
$23\times 2+12+3\times 16+18x=141.41$
$\therefore x\cong 2$