Question

The $8.9\mathrm{M}$ ${\mathrm{H}}_2{\mathrm{O}}_2$at $273\mathrm{K}$ and $1\mathrm{atm}$ its volume strength will be$—$$\left(\mathrm{R}=0.0821{\mathrm{LatmK}}^{-1}{\mathrm{mol}}^{-1}\right)$(Round it off to the nearest integer)

Answer 1

Step 1:

$2{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+1{\mathrm{O}}_2\left(\mathrm{g}\right)\uparrow$

Step 2:

$2$moles of ${\mathrm{H}}_2{\mathrm{O}}_2$$=22.4\mathrm{l}$

$1$moles of ${\mathrm{H}}_2{\mathrm{O}}_2$$=11.2\mathrm{l}$

$\begin{array}{rcl}\mathrm{Volume}\mathrm{strength}& =& 11.2\times \mathrm{strength}\\ & =& 11.2\times 8.9\\ & =& 99.68\cong 100\end{array}$

Therefore, the volume strength of ${\mathrm{H}}_2{\mathrm{O}}_2$ at $273\mathrm{K}$and $1$ atm is $100$.