A) 0.043
B) 4.4
C) 3.4
D) 3.86
E) 0.437
Correct Answer: E
Solution :
No. of molecules \[=2\times V.d\] \[2\times 38.3=76.3\] wt. of \[N{{O}_{2}}=x\] So that wt. of \[{{N}_{2}}{{O}_{4}}=100-x\] Hence, \[\frac{x}{46}+\frac{100-x}{92}=\frac{100}{76.6}=\frac{2x+100-x}{92}=\frac{100}{76.6}\] x = 20.10, no. of mole. of \[N{{O}_{2}}=\frac{20.10}{46}=0.437\]You need to login to perform this action.
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