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