The mole percentage of oxygen in a mixture of 7.0 g of nitrogen and 8.0 g of oxygen is

Options

A. 8
B. 16
C. 24
D. 50

Answer

d

Explanation

No. of moles of `N_(2)= 7/28 = 0.25`
No. of moles of `O_(2) = 8/32 = 0.25`
`therefore` Mole percentage of `O_(2)`
((moles of `O_(2)`)/( Moles of `N_(2)` + Moles of `O_(2)`)) `xx 100= (0.25/(0.25+0.25)) xx 100= 50%`