91Ó°ÊÓ

Given, the volume of oxygen gas is \(57.2\, \mathrm{mL}\), the temperature is \(22^{\circ} \mathrm{C}\) which is equal to \((22+273) \, \mathrm{K}\), and the total pressure is \(734 \, \mathrm{torr}\). The vapor pressure of water at this temperature is \(19.8 \, \mathrm{torr}\). Let's find out the pressure exerted by oxygen gas alone: $$ P_\mathrm{O_2}= P_\mathrm{total} - P_\mathrm{water} = 734 - 19.8 = 714.2 \, \mathrm{torr} $$ Now we can convert the pressure to atm: $$ P_\mathrm{O_2} = 714.2 \, \mathrm{torr} \times \frac{1 \, \mathrm{atm}}{760 \, \mathrm{torr}} = 0.9403 \, \mathrm{atm} $$ Next, we need to convert the volume of gas from mL to L: $$ V_\mathrm{O_2} = 57.2 \, \mathrm{mL} \times \frac{1 \, \mathrm{L}}{1000 \, \mathrm{mL}} = 0.0572 \, \mathrm{L} $$ The ideal gas law is given by \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Rearranging to find n, we have: $$ n_\mathrm{O_2} = \frac{PV}{RT} $$ Substitute the given values (using \(R = 0.0821 \, \mathrm{L \cdot atm \cdot mol^{-1} \cdot K^{-1}}\)) and find the moles of oxygen gas: $$ n_\mathrm{O_2} = \frac{(0.9403 \, \mathrm{atm})(0.0572 \, \mathrm{L})}{(0.0821 \, \mathrm{L \cdot atm \cdot mol^{-1} \cdot K^{-1}})(295 \, \mathrm{K})} = 0.00227 \, \mathrm{mol} $$

The given balanced chemical equation for the decomposition of sodium chlorate is: $$ 2\mathrm{NaClO_3} \rightarrow 2\mathrm{NaCl} + 3\mathrm{O_2} $$ Using stoichiometry, we can find the moles of sodium chlorate that decomposed. $$ n_\mathrm{NaClO_3} = \frac{2}{3} n_\mathrm{O_2} = \frac{2}{3}(0.00227 \, \mathrm{mol}) = 0.001514 \, \mathrm{mol} $$

The molar mass of sodium chlorate (\(\mathrm{NaClO}_{3}\)) is approximately \(106.44 \, \mathrm{g \cdot mol^{-1}}\). Now we can find the mass of sodium chlorate (\(\mathrm{NaClO}_{3}\)) that decomposed. $$ m_\mathrm{NaClO_3} = n_\mathrm{NaClO_3} \times \mathrm{Molar \, mass} = (0.001514 \, \mathrm{mol})(106.44 \, \mathrm{g \cdot mol^{-1}}) = 0.1611 \, \mathrm{g} $$

The mass percent of sodium chlorate in the original sample can be calculated using the formula: $$ \mathrm{Mass \, percent} = \frac{\mathrm{Mass \, of \, decomposed \, NaClO_3}}{\mathrm{Mass \, of \, original \, sample}} \times 100 $$ Substituting the given values: $$ \mathrm{Mass \, percent} = \frac{0.1611 \, \mathrm{g}}{0.8765 \, \mathrm{g}} \times 100 = 18.37\% $$ So, the mass percent of sodium chlorate (\(\mathrm{NaClO}_{3}\)) in the original sample is approximately \(18.37 \%\).