A) \[7.47L\]
B) \[0.33L\]
C) \[44.8L\]
D) \[23.5L\]
Correct Answer: A
Solution :
| surface area of 1 g carbon \[=1000{{\operatorname{m}}^{2}}\] |
| \[={{10}^{7}}c{{\operatorname{m}}^{2}}\] |
| Total surface area of carbon |
| \[=\frac{44}{7}\times {{10}^{7}}{{\operatorname{cm}}^{2}}\] |
| No. of \[{{\operatorname{NH}}_{3}}\] molecules adsorbed \[=\frac{\frac{22}{7}\times {{10}^{7}}}{\frac{22}{7}\times {{10}^{-16}}}\times {{\operatorname{N}}_{A}}\] |
| Vol. of \[{{\operatorname{NH}}_{3}}\]adsorbed at atm 273K\[=\frac{2\times {{10}^{23}}}{6\times {{10}^{23}}}\times 22.4\] |
| \[=7.47L\] |
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