Question

$0.7$ grams of $Na _{2} CO _{3} \cdot x H _{2} O$ 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 Report your answer by rounding it upto nearest whole number.

Answer 1

Meq. of $Na _{2} CO _{3} \cdot xH _{2} O$ in $20 mL =19.8\times \frac{1}{10}$
$\therefore $ Meq. of $Na _{2} CO _{3} \cdot xH _{2} O Na _{2} CO _{3} \cdot xH _{2} O$ in $100\, mL =19.8 \times \frac{1}{10} \times 5$
$\therefore \frac{w}{E} \times 1000=19.8 \times \frac{1}{10} \times 5$
or $\frac{0.7}{\frac{M}{2}} \times 1000=\frac{19.8}{2}$
$\therefore M=141.41$
$23 \times 2+12+3 \times 16+18 x=141.41$
$\therefore x \cong 2$