A) \[0.22\]
B) \[0.44\]
C) \[~0.66\]
D) \[0.88\]
Correct Answer: D
Solution :
\[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{p}_{1}}}{{{p}_{2}}}\sqrt{\frac{{{M}_{2}}}{{{M}_{1}}}}\] \[{{P}_{He}}={{x}_{He}}\cdot {{p}_{total}}\] \[=\frac{1}{4}\times 16=4\,\,atom\] \[{{P}_{{{N}_{2}}}}={{x}_{{{N}_{2}}}}\cdot {{p}_{total}}\] \[=\frac{3}{4}\times 16=12\,\,atom=(p_{total}^{-}{{p}_{He}})\] \[\frac{{{r}_{He}}}{{{r}_{{{N}_{2}}}}}=\frac{{{p}_{He}}}{{{p}_{{{N}_{2}}}}}\sqrt{\frac{M({{N}_{2}})}{M(He)}}\] \[=\frac{4}{12}\sqrt{\frac{28}{4}}=0.88\] Hence, moles of\[He\]and\[{{N}_{3}}\]effusing out initially are in the ratio\[0.88:1\].
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