A) 10 vol.strength = 0.893M = 1.786N = 3.03% (w/v)
10 vol.strength ⇒ 3.03% (w/v)
50 vol.strength ⇒ ??

$$\begin{gathered} \frac{50\times 3.03}{10}=15.15\%(\mathrm{w}/\mathrm{v}) \\ \mathrm{M}= \frac{\mathrm{Volume} \mathrm{strength}}{11.2} \\ \mathrm{M}=\frac{50}{11.2}=4.46\mathrm{M} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{solution} \end{gathered}$$

B) 1 volume strength of H2O2
10 vol = 3.03% (w/v)
1 vol = ??
 

 

 

C) 30%(w/v) of H2O2 means
 

$$\begin{gathered} 30 \mathrm{gr} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \stackrel{\mathrm{present} \mathrm{in} }{\to }100 \mathrm{ml} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{solution}(0.1\mathrm{L}) \\ ? \mathrm{gr} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2\stackrel{\mathrm{present} \mathrm{in} }{\to } 1000\mathrm{ml} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{solution}\left(1\mathrm{L}\right) \\ \frac{ 1\times 30}{0.1}=300 \mathrm{gr} \end{gathered}$$

i.e.. 300gr/liter
D) 4 MOlar H2O2 solution
10vol = 0.893M
?? vol = 4M

$\frac{4\times 10}{0.893}=44.79\mathrm{vol} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2$