Question

$10$ mL of $0.1$ M $H_{2}S{O}_4$ is needed for complete the reaction with $25$ mL of $11.44$ g/L $N{a}_{2}C{O}_3.x{H}_{2}O$. The value of $x$ is:

• A. $0$
• B. $2$
• C. $6$
• D. $10$

Answer 1

The correct option is B $2$

eq. of $H_{2}S{O}_4=$ eq. of $N{a}_{2}C{O}_3.x{H}_{2}O$

$10\times 0.1\times 2=25\times N$

$N=\frac{20\times 0.1}{25}\Rightarrow N=\frac{2}{25}$

$=0.08=normally$

here strength = $N\times eq.wt$

$11.44=0.08\times eq.wt$

Eq. wt $=143=46+12+48+18x$

$106+18x=143$

$[\therefore$ Eq. wt = mol. wt.]

$x=2.05$