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The maximum number of molecules is present in :A. 15 L of `H_(2)` gas at STPB. 5 L of `N_(2)` gas at STPC. 0.5 g of `H_(2)` gasD. 10 g of `O_(2)` gas |
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Answer» Correct Answer - A In 15 L of `H_(2)` gas at STP, the number of molecules `=(6.023xx10^(23))/(22.4)xx15` `=4033xx10^(23)` In 5 L of `N_(2)` gas at STP, ltbr. the number of molecules `=(6.023xx10^(23)xx5)/(22.4)` `=1.344xx10^(23)` In 0.5 g of `H_(2)` gas, the number of molecules `= (6.23xx10^(23)xx0.5)/(2)` `=1.505xx10^(23)` In 10 g of `O_(2)` gas , the number of molecules `=(6.23xx10^(23)xx10)/(32)` `=1.882xx10^(23)` Hence , maximum number of molecules are presents in 15 L of `H_(2)` at STP. |
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