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The maximum number of molecules is present inA. 10 g of `O_(2)` gasB. 15L of `H_(2)` gas at S.T.P.C. 5L of `N_(2)` gas at S.T.P.D. 0.5 g of `H_(2)` gas |
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Answer» Correct Answer - B More the volume, more is the molecules. Converted the choices (A) and (D) into volumes. (A) 32 g of `O_(2)` at S.T.P = 22.4 L `:.` 10 g `O_(2)` at S.T.P. `= (22.4)/(32)xx10L=7L` (Molar mass of `O_(2) = 32 g mol^(-1)`) (D) 2 g of `H_(2)` at S.T.P. = 22.4 L 0.5 g of `H_(2)` at S.T.P. `= (22.4)/(2)xx0.5L=5.6L` Thus we have 7L of `O_(2), 15 L` of `H_(2)`, 5 L of `N_(2)` and 5.6 L of `H_(2)` (at S.T.P.) Clearly 15 L of `H_(2)` has maximum number of molecules |
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