A) 7
B) 5
C) 4
D) 3
Correct Answer: C
Solution :
Key Idea Molecular mass = 2 \[\times \] vapour density Molecular mass = n \[\times \] empirical mass Here, Vapour density = 56 \[\therefore \] Molecular mass \[=2\times 56=112\] Empirical mass of \[{{(CO)}_{n}}=(1\times 12+1\times 16)\] = 28 \[n=\frac{molecular\text{ }mass}{empirical\text{ }mass}=\frac{112}{28}\]You need to login to perform this action.
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