A) \[125mL/g\]
B) \[16.25mL/g\]
C) \[25mL/g\]
D) None of these
Correct Answer: B
Solution :
Final volume of gas, at 608 torr pressure \[{{\operatorname{V}}_{2}}=\frac{{{P}_{1}}{{V}_{1}}}{{{P}_{2}}};=\frac{760\times 1}{608}\]\[\Rightarrow \]\[1.25L\] or \[{{V}_{2}}=1250mL\] |
Volume occupied by gas =volume of vessel -volume occupied by charcoal \[=1000-\frac{16}{1.6}=990\operatorname{mL}\] |
Difference of volume is due to adsorption of gas by charcoal |
\[\therefore \]Volume of gas adsorbed by charcoal \[=1250-990\]\[\Rightarrow \]\[260\operatorname{mL}\] |
Volume of the gas adsorbed per gram of charcoal |
\[=\frac{260}{16}=16.25\operatorname{mL}/g\,\]at 608 torr and \[27{}^\circ \operatorname{C}.\] |
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