Question
What will be the strength of $$20\,vol$$ of $${H_2}{O_2}$$ in terms of gram per litre?
A.
$$60.71\,g\,{L^{ - 1}}$$
B.
$$5.6\,g\,{L^{ - 1}}$$
C.
$$30.62\,g\,{L^{ - 1}}$$
D.
$$17\,g\,{L^{ - 1}}$$
Answer :
$$60.71\,g\,{L^{ - 1}}$$
Solution :
$$20\,vol$$ of $${H_2}{O_2}$$ means :
$$1\,L$$ of this $${H_2}{O_2}$$ will give $$20\,L$$ of oxygen at $$STP.$$
$$\mathop {2{H_2}{O_2}}\limits_{2 \times 34} \,\,\,\,\, \to \mathop {{O_2}}\limits_{22.4\,L\,\,{\text{at}}\,\,STP} + {H_2}O$$
$$22.4\,L$$ of $${{O_2}}$$ is produced from $$68\,g$$ of $${H_2}{O_2}$$
$$20\,L$$ of $${{O_2}}$$ is produced from $$ = \frac{{68 \times 20}}{{22.4}} = 60.71\,g\,{L^{ - 1}}$$ of $${H_2}{O_2}.$$