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Q.
The maximum number of molecules is present in
Bihar CECEBihar CECE 2008Some Basic Concepts of Chemistry
Solution:
In $15\, L$ of $H _{2}$ gas at $STP$,
the number of molecules $=\frac{6.023 \times 10^{23}}{22.4} \times 15$
$=4.033 \times 10^{23}$
In $5\, L$ of $N _{2}$ gas at $STP$,
the number of molecules $=\frac{6.023 \times 10^{22-} \times 5}{22.4} $
$=1.344 \times 10^{23}$
In $0.5 \,g$ of $H _{2}$ gas,
the number of molecules $=\frac{6.023 \times 10^{23} \times 0.5}{2}$
$=1.505 \times 10^{23}$
In $10\, g$ of $O _{2}$ gas,
the number of molecules $=\frac{6.023 \times 10^{23} \times 10}{32} $
$=1.882 \times 10^{23}$
Hence, maximum number of molecules are present in $15\, L$ of $H _{2}$ at STP.