A) \[11/20\text{ }g\]
B) \[20/11\text{ }g\]
C) \[5/63\text{ }g\]
D) \[63/5\text{ }g\]
Correct Answer: D
Solution :
[d] \[PV=\frac{m}{M}RT\] \[P\propto m.T\] \[\frac{10}{5}=\frac{28}{m}\times \frac{330}{300}\] \[\Rightarrow \,\,\,\,m=\frac{28\times 330\times 5}{10\times 300}=\frac{231}{15}gm=\frac{77}{5}gm\] Gas leaked \[=28-\frac{77}{5}=\frac{63}{5}gm\]You need to login to perform this action.
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