A) \[0.032\text{ }kg\text{ }mo{{l}^{-1}}\]
B) \[0.016\text{ }kg\text{ }mo{{l}^{-1}}\]
C) \[0.004\text{ }kg\text{ }mo{{l}^{-1}}\]
D) \[0.008\text{ }kg\text{ }mo{{l}^{-1}}\]
Correct Answer: D
Solution :
: Ace. to Grahams law of diffusion. \[\frac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\frac{{{M}_{2}}}{{{M}_{1}}}}\] \[\frac{{{r}_{1}}}{2{{r}_{1}}}=\sqrt{\frac{{{M}_{2}}}{32}}\] \[\frac{1}{4}=\sqrt{\frac{{{M}_{2}}}{32}};\] \[{{M}_{2}}=8\]You need to login to perform this action.
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