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Maximum number of molecules are present inA. 15 L of `H_(2)` gas at S.T.PB. 5 L of `N_(2)` gas at S.T.PC. 0.5 of `H_(2)` gasD. 10 g of `O_(2)` gas |
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Answer» Correct Answer - A (a) 15 L of `H_(2)` gas at S.T.P `= (15L)/((22.4L))xx6.022xx10^(23)=4.033xx10^(23)` (b) 5L of `N_(2)` gas at S.T.P. `= ((5L))/((22.4L))xx6.022xx10^(23)` `=1.344xx10^(23)` (c) 0.5 g of `H_(2)=((0.5g))/((2.0g))xx6.022 xx 10^(23)` `=1.505 xx 10^(23)` 10 g of `O_(2)=((10.0g))/((32.0g))xx6.022xx10^(23)` `=1.882xx10^(23)`. |
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