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21.The maximum number of molecules are presentIAIPMT 2004]In(A) 15 L of N2 gas at STP(B) 5L of N2 gas at STP(C) 0.5 g of H2 gas(D) 10 g of 02 gas |
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Answer» (i) Number of moles in 15 L of N2gas at STP. = (15 L / 22.4 L) = 0.67 moles (ii)Number of moles in 5 L of N2gas atSTP. = (5 L / 22.4 L) = 0.22 moles (iii) Number of moles in0.5 g of H2gas = (0.5 g / 2 g mol-1) = 0.25 moles (iv) Number of moles in10 g of O2gas = (10 g / 32gmol-1) = 0.312 moles Hence, as we know that 1 mole of any gas has Avogadro number of molecules (6.022 x 1023) in it, we can say that the higher number of moles i.e. in 15 L of N2gas will contain maximum number of molecules, as it has higher number of molar quantity. |
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