A) 95.39
B) 39.95
C) 99.53
D) 59.93
Correct Answer: D
Solution :
[d] \[\text{V}\text{.D}\text{.}=\frac{Wt.\,of\,45ml.\,of\,vapours\text{ }at\text{ }NTP}{Wt.of\,45ml.of\,{{H}_{2}}\,at\,NTP}\] \[=\frac{0.24g}{45ml.\times 0.000089gm{{l}^{-1}}}=59.93\]You need to login to perform this action.
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