A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 mL of $$CO_2$$ at T = 298.15 K and P = 1 bar. If molar volume of $$CO_2$$ is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet? [Molar mass of $$NaHCO_3 = 84$$ g $$mol^{-1}$$]

We have been told that the tablet liberates $$0.25\;{\rm mL}$$ of carbon dioxide when it reacts. The molar volume of the gas at the stated temperature and pressure is given as $$25.0\;{\rm L}$$ per mole.

First we put both volumes in the same unit so that the division is straightforward. Remembering that

$$1\;{\rm L}=1000\;{\rm mL},$$

we write

$$25.0\;{\rm L}=25.0\times1000\;{\rm mL}=25\,000\;{\rm mL}.$$

The definition of molar volume is

$$V_{\rm m}=\frac{V}{n},$$

so rearranging we obtain the number of moles:

$$n=\frac{V}{V_{\rm m}}.$$

Substituting the numerical values,

$$n_{CO_2}=\frac{0.25\;{\rm mL}}{25\,000\;{\rm mL\,mol^{-1}}} =\frac{0.25}{25\,000}\;{\rm mol} =1.0\times10^{-5}\;{\rm mol}.$$

In the effervescent reaction each mole of sodium bicarbonate produces one mole of carbon dioxide, because the neutralisation of $$NaHCO_3$$ by an acid is represented by the general step

$$NaHCO_3+H^+\rightarrow Na^++CO_2+H_2O,$$

so the stoichiometric ratio is $$1:1.$$ Therefore

$$n_{NaHCO_3}=n_{CO_2}=1.0\times10^{-5}\;{\rm mol}.$$

Next we change moles of bicarbonate to mass by using its molar mass. The relationship is

$$m=nM.$$

Substituting,

$$m_{NaHCO_3}=(1.0\times10^{-5}\;{\rm mol})(84\;{\rm g\,mol^{-1}}) =8.4\times10^{-4}\;{\rm g}.$$

Converting grams to milligrams ( $$1\;{\rm g}=1000\;{\rm mg}$$ ),

$$m_{NaHCO_3}=8.4\times10^{-4}\;{\rm g}\times1000 =0.84\;{\rm mg}.$$

The tablet itself has a mass of $$10\;{\rm mg}.$$ The percentage of sodium bicarbonate present is therefore

$$\%\;NaHCO_3=\frac{0.84\;{\rm mg}}{10\;{\rm mg}}\times100 =8.4\%.$$

Hence, the correct answer is Option D.