Question

The volume of oxygen gas liberated at $ 27{{\,}^{o}}C $ and 760 mm Hg pressure when 24.5 g of $ \text{KCl}{{\text{O}}_{\text{3}}} $ is heated, is $ \text{(M}\text{.wt}\text{.of KC1}{{\text{O}}_{\text{3}}}\text{=122}\text{.5)} $

• A. 4.48 L
• B. 9.6 L
• C. 7.4 L
• D. 11.2 L

Answer 1

Answer: (C)

$ \begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underset{\begin{smallmatrix} 2\times 122.5 \\ =\,245.0 \end{smallmatrix}}{\mathop{KCl{{O}_{3}}}}\,\xrightarrow{{}}\underset{\begin{smallmatrix} 3\,\,\times \,22.4 \\ =\,\,67.2L \end{smallmatrix}}{\mathop{2KCl+3{{O}_{2}}}}\, \\ & at\,STP\,\,\,\,\,\,\,\,\, \\ \end{align} $ $ \because $ When $ 245\,g\,KCl{{O}_{3}} $ is heated, liberated oxygen = 67.2 L $ \therefore $ When $ 24.5\text{ }g\text{ }KCl{{O}_{3}} $ is heated, liberated oxygen $ =\frac{67.2\times 24.5}{245}=6.72\,L $ This volume of oxygen was liberated at STP. Therefore, at $ {{27}^{o}}C $ and 760 mm Hg pressure, the volume of liberated oxygen will be (STP) (Given conditions) $ \frac{{{p}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{p}_{2}}{{V}_{2}}}{{{T}_{2}}} $ $ \frac{760\times 6.72}{273}=\frac{760\times {{V}_{2}}}{300} $ $ \therefore $ $ {{V}_{2}}=\frac{760\times 6.72\times 300}{760\times 273} $ $ \text{=}\,\text{7}\text{.38}\,\text{L} $ $ \approx \,\text{7}\text{.4}\,\text{L} $