1.

The maximum number of molecules is present in

Answer»

15 L of `H_(2)` gas at STP
5L of `N_(2)` gas at STP
0.5 g of `H_(2)` gas
10 g of `O_(2)` gas

Solution :At STP, 22.4 L of any gas `=6.02 xx 10^(23)` molecules
`therefore 15 L H_(2) = (6.02 xx 10^(23))/(22.4) xx 15 = 4.03 xx 10^(23)` molecules
`5 L N_(2) = (6.02 xx 10^(23))/22.4 xx 5 = 1.344 xx 10^(23)` molecules
2g `H_(2) = 6.02 xx 10^(23)` molecules
`therefore 0.5 g H_(2) = (6.02 xx 10^(23))/2 xx 0.5` molecules
`=1.505 xx 10^(23)` molecules
32 g of `O_(2) = 6.02 xx 10^(23)` molecules
`therefore 10 g O_(2) = (6.02 xx 10^(23))/32 xx 10` molecules
`=1.88 xx 10^(23)` molecules.


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