A) 6.4 x102
B) 3.0 x102
C) 1.6 x105
D) 5.6x104
Correct Answer: C
Solution :
\[{{W}_{2}}=4g.V=1000mL=1L,\pi =6.0\times {{10}^{-4}}atm\] \[T=300K,R=00.0821Latm,{{K}^{-1}}mo{{l}^{-1}}\] \[\pi ={{W}_{2}}\times RT/({{M}_{2}}V)\] \[\therefore \]\[{{M}_{2}}=4\times 0.0821\times 300/(6\times {{10}^{-14}}\times 1)=16.42\times {{10}^{4}}\]\[=1.6\times {{10}^{5}}\]
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