The volume strength of 8.9 M $$H_2O_2$$ solution calculated at 273 K and 1 atm is ......... ($$R = 0.0821$$ L atm K$$^{-1}$$ mol$$^{-1}$$) (rounded off to the nearest integer)

First, recall the definition of volume strength. It is the volume (expressed in millilitres, at 273 K and 1 atm) of $$O_2$$ gas that is liberated from 1 mL of a hydrogen-peroxide solution when the peroxide decomposes completely.

The balanced decomposition reaction is

$$2\,H_2O_2 \;\longrightarrow\; 2\,H_2O + O_2$$

From this equation, 2 moles of $$H_2O_2$$ give 1 mole of $$O_2$$.

We are given a solution that is $$8.9\;{\rm M}$$, meaning

$$8.9\;{\rm mol\;H_2O_2\;per\;L\;of\;solution}.$$

So, in exactly $$1\;{\rm L} = 1000\;{\rm mL}$$ of this solution, the moles of peroxide are

$$n_{\!H_2O_2}=8.9.$$

Using the stoichiometric ratio $$\dfrac{1\;{\rm mol}\;O_2}{2\;{\rm mol}\;H_2O_2},$$ the moles of oxygen that can be produced from these 8.9 moles of peroxide are

$$n_{O_2}= \dfrac{8.9}{2}=4.45\;{\rm mol}.$$

To find the volume occupied by these $$O_2$$ moles at 273 K and 1 atm, we invoke the ideal-gas equation

$$PV = nRT.$$

Substituting $$P = 1\;{\rm atm},\; n = 4.45\;{\rm mol},\; R = 0.0821\;{\rm L\,atm\,K^{-1}\,mol^{-1}},\; T = 273\;{\rm K},$$ we obtain

$$V = \dfrac{nRT}{P} = 4.45 \times 0.0821 \times 273.$$

Now multiply step by step:

$$0.0821 \times 273 = 22.4133,$$ $$4.45 \times 22.4133 \approx 99.7392\;{\rm L}.$$

Thus, $$1\;{\rm L}$$ of the solution liberates approximately $$99.74\;{\rm L}$$ of $$O_2$$ gas at the stated conditions.

Convert this gas volume to millilitres because volume strength is stated per millilitre of solution:

$$99.74\;{\rm L} = 99.74 \times 1000 = 9.974 \times 10^{4}\;{\rm mL}.$$

Since these $$9.974 \times 10^{4}\;{\rm mL}$$ of gas come from $$1000\;{\rm mL}$$ of solution, the volume of gas corresponding to 1 mL of solution is

$$\dfrac{9.974 \times 10^{4}\;{\rm mL}}{1000\;{\rm mL}} = 99.74\;{\rm mL}.$$

Rounded to the nearest whole number, this becomes $$100\;{\rm mL}.$$

Therefore, the solution is called a 100-volume hydrogen-peroxide solution.

Hence, the correct answer is Option A (100).