A) 5.0246
B) 4.1246
C) 3.0196
D) 5.1969
Correct Answer: D
Solution :
\[d=\frac{pM}{RT}\] \[=\frac{2\text{ }atm\times 64\text{ }g\text{ }mo{{l}^{-1}}}{0.0821\text{ }L\text{ }atm\text{ }{{K}^{-1}}\text{ }mo{{l}^{-1}}\times 300\text{ }K}\] \[=5.1969\text{ }g{{L}^{-1}}\]You need to login to perform this action.
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