1.

If 15mg of N_(2)O_(3) is added to 4.82 xx 10^(20) molecules of N_(2)O_(3), thetotal volume occupied by thegas at STP is

Answer»

`0.044` L
` 0.022` L
` 0.22` L
` 0.44` L

Solution :Gram MOLECULAR weight of `N_(2)O_(3)` is 76 g AND76 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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