The strengths of 5.6 volume hydrogen peroxide (of density 1 g/mL) in terms of mass percentage and molarity(M) respectively, are: (Take molar mass of hydrogen peroxide as 34 g/mol)

First, we recall the definition of “volume strength”. An $$x$$-volume hydrogen peroxide solution is one in which 1 volume of the solution releases $$x$$ volumes of $$\mathrm{O_2}$$ gas at NTP (STP). Here the given solution is “5.6-volume”, so

$$1\;\text{mL solution}\;\longrightarrow\;5.6\;\text{mL } \mathrm{O_2}\text{(g at NTP)}$$

We now convert this released oxygen to moles. At NTP,

$$\text{Molar volume of any gas}=22.4\;\text{L}=22.4\times10^{3}\;\text{mL}$$

So, moles of $$\mathrm{O_2}$$ obtained from 1 mL solution are

$$n(\mathrm{O_2})=\frac{5.6\;\text{mL}}{22.4\times10^{3}\;\text{mL mol}^{-1}} =\frac{5.6}{22.4}\times10^{-3}\;\text{mol} =0.25\times10^{-3}\;\text{mol} =2.5\times10^{-4}\;\text{mol}$$

The decomposition reaction of hydrogen peroxide is

$$2\,\mathrm{H_2O_2}\;\longrightarrow\;2\,\mathrm{H_2O}+ \mathrm{O_2}$$

From the stoichiometry, 2 mol $$\mathrm{H_2O_2}$$ give 1 mol $$\mathrm{O_2}$$, hence

$$n(\mathrm{H_2O_2}) = 2 \times n(\mathrm{O_2}) = 2 \times 2.5\times10^{-4}\;\text{mol} = 5.0\times10^{-4}\;\text{mol}$$

This amount of $$\mathrm{H_2O_2}$$ is present in 1 mL of the solution. We now express the concentration in molarity, which is moles per litre (1000 mL).

$$\text{Molarity},\;M =\frac{5.0\times10^{-4}\;\text{mol in 1 mL}}{1\;\text{mL}} \times 1000\;\frac{\text{mL}}{\text{L}} = 0.50\;\text{mol L}^{-1}$$

Thus the solution is $$0.5\;M$$.

Next we calculate the mass percentage. The moles of $$\mathrm{H_2O_2}$$ in 1 L (1000 mL) of solution are 0.50 mol, so the mass of peroxide present is

$$m(\mathrm{H_2O_2}) = 0.50\;\text{mol}\times34\;\text{g mol}^{-1} = 17\;\text{g}$$

Because the density is 1 g mL-1, 1 L of solution has a mass of 1000 g. Therefore,

$$\text{Mass percentage} = \frac{17\;\text{g}}{1000\;\text{g}}\times100 = 1.7\%$$

We have now obtained both required strengths: 1.7 % (w/w) and 0.5 M.

Hence, the correct answer is Option A.