A) 4.48 L
B) 9.6 L
C) 7.4 L
D) 11.2 L
Correct Answer: C
Solution :
\[\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underset{\begin{smallmatrix} 2\times 122.5 \\ =\,245.0 \end{smallmatrix}}{\mathop{KCl{{O}_{3}}}}\,\xrightarrow{{}}\underset{\begin{smallmatrix} 3\,\,\times \,22.4 \\ =\,\,67.2L \end{smallmatrix}}{\mathop{2KCl+3{{O}_{2}}}}\, \\ & at\,STP\,\,\,\,\,\,\,\,\, \\ \end{align}\] \[\because \]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) \[\frac{{{p}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{p}_{2}}{{V}_{2}}}{{{T}_{2}}}\] \[\frac{760\times 6.72}{273}=\frac{760\times {{V}_{2}}}{300}\] \[\therefore \] \[{{V}_{2}}=\frac{760\times 6.72\times 300}{760\times 273}\] \[\text{=}\,\text{7}\text{.38}\,\text{L}\]\[\approx \,\text{7}\text{.4}\,\text{L}\]
You need to login to perform this action.
You will be redirected in
3 sec
