A) \[15L\,\]of\[{{H}_{2}}\]gas at STP
B) \[5\,\,\,L\]of\[{{\text{N}}_{\text{2}}}\] gas at STP
C) \[\text{0}\text{.5}\]g of \[{{\text{H}}_{2}}\]gas
D) \[10g\,\]of\[{{O}_{2}}\]
Correct Answer: A
Solution :
In 15 L of Ha gas at STP, the number of molecules \[=\frac{6.023\times {{10}^{23}}}{22.4}\times 15\] \[=4.033\times {{10}^{23}}\] In \[\text{5 L}\]of \[{{\text{N}}_{\text{2}}}\]gas at STP \[=\frac{\text{6}\text{.023}\times \text{1}{{\text{0}}^{23}}\times 5}{22.4}=1.344\times {{10}^{23}}\] In \[0.5g\]of \[{{H}_{2}}\]gas \[\frac{6.023\times {{10}^{23}}\times 0.5}{2}\] \[=1.505\times {{10}^{23}}\] In \[10g\]of\[{{\text{O}}_{2}}\]gas \[=\frac{6.023\times {{10}^{23}}}{32}=1.882\times {{10}^{23}}\] Hence, maximum molecules are present in \[15\,\text{L}\]of\[{{H}_{2}}\]at STP.
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