Question

What weight of $H_2$ NTP could be contained in a vessel that holds 4.8 gm $O_2$ at NTP?

• A. 4.9 gm
• B. 0.3 gm
• C. 0.6 gm
• D. 3.5 gm

Answer 1

The correct option is B 0.3 gm

Solution:-

Let $W$ be the weight of $H_2$.

According to Avogadro's law, equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

Therefore,

No. of molecules of $H_2=$ No. of molecules of $O_2$

$\frac{W\times N_a}{2}=\frac{4.8\times N_a}{32}$

$\Rightarrow W=\frac{2\times 4.8}{32}=0.3gm$

Hence the weight of NTP could be $0.3gm$.