A) 0.90
B) 1.11
C) 0.11
D) 1.0
Correct Answer: A
Solution :
\[\frac{{{P}_{1}}{{V}_{1}}}{{{n}_{1}}{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{n}_{2}}{{T}_{2}}}\therefore {{n}_{2}}=\frac{{{P}_{2}}{{V}_{2}}{{T}_{1}}}{{{P}_{1}}{{V}_{1}}{{T}_{2}}}{{n}_{1}}\] \[\left. \begin{align} & at\text{ }STPone\text{ }mole. \\ & {{P}_{1}}=\text{ }1atm. \\ & {{V}_{1}}=\text{ }22.4lt \\ & {{T}_{1}}=\text{ }{{273}^{o}}K \\ \end{align} \right\}\begin{matrix} at\,T={{273}^{o}}+{{30}^{o}}={{303}^{o}}K \\ {{P}_{2}}=1atm \\ {{V}_{2}}=22.4 \\ \end{matrix}\] \[{{n}_{2}}=\frac{1}{1}\times \frac{22.4}{22.4}\times \frac{273}{303}\times 1=0.9\]moles
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