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If 15 mg of `N_(2)O_(3)` is added to `4.82 xx 10^(20)` molecules of `N_(2)O_(3)`, the total volume occupied by the gas at STP isA. `0.044` LB. ` 0.022` LC. ` 0.22` LD. ` 0.44` L |
Answer» Gram molecular weight of `N_(2)O_(3)` is 76 g and 76 g of `N_(2)O_(3)` contains `6.023 xx 10^(23)` molecules ` = (76 xx 4.82 xx 10^(20))/(6.023 xx 10^(23)) g = 0.061` g 15 mg of ` N_(2)O_(3) = 0.015` g `:." Total weight of "N_(2)O_(3) = 0.061 + 0.015 = 0.076` g 76 g occupies 22.4 L at STP. 0.076 g occupies ? L at STP. ` = (0.076 xx 22.4)/(76) = 0.022` L |
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