Question

Match the List I with List II and choose the correct option from the codes given below.

Answer 1

(a) 10 Volumes of $H_2{O}_2$ :

$1N$ of $H_2{O}_2⟶5.6V$

$5.6V\stackrel{gives}{⟶}$ ${1N\text{of}H}_2{O}_2$

$1V\stackrel{gives}{⟶}\frac{1}{5.6}\text{of}{H}_2{O}_2$

$10V\stackrel{gives}{⟶}10\times \frac{1}{5.6}\text{of}{H}_2{O}_2$

Now we kuow

percentage Strength of $1N{H}_2{O}_2$ solution

$=1.7\%$

for $\frac{10}{5.6}\times 1$ the $\%$ strength $=\frac{1.7\times 10}{5.6}=\text{3.03\%}$

b) 20 volumes of $H_{2}0:-$

$1V$ of $H_2{O}_2$ solution gries $20V$ of $O_2$

$1ml$ of $H_2{O}_2$ solution gives $20ml$ of $O_2$

volume slrength $=11.2\times$ molarity

$20V=11.2\times M$

$\frac{20}{11.2}M\Rightarrow 1.785\text{M}$

c) $30V$ of $H_2{O}_2$ : $-$

volume strength $=5.6\times N$

$30V=5.6\times N$

$\frac{30V}{5.6}=N$

$\text{5.358 N}$

(d) $100V$ of $H_{2}O$ is known as perhydral having $30\%$ w/v of $H_2{O}_2$.

$\begin{array}{cc}\text{Option:}& A-4\\ B& -3\\ C& -2\\ D& -1\end{array}$