Question

2 g of hydrogen will have same number of molecules as is present in:

• A. 14 gm of nitrogen
• B. 32 gm of $S{O}_2$
• C. 71 gm of chlorine
• D. 24 gm of ozone

Answer 1

The correct option is C 71 gm of chlorine

No. of molecules in 2g of hydrogen $=N_A$
A) No. of molecules in 14g of nitrogen $=\frac{14\times N_A}{28}=0.5N_A$
B) No. of molecules in 32g of $S{O}_2$ $=\frac{32\times N_A}{64}=0.5N_A$
C) No. of molecules in 71g of $C{l}_2$ $=N_A$
D) No. of molecules in 24g of ozone $=\frac{24\times N_A}{48}=0.5N_A$
Therefore 71g of $C{l}_2$ has same no. of molecules as in 2g of hydrogen.