1.

The number of molecules in 4.25 g of ammonia are approximately

Answer»

`0.5 XX 10^23`
`1.5 xx 10^23`
`2.5 xx 10^23`
`3.5 xx 10^23`

SOLUTION :No. of MOLECULES of `NH_3` in 17G ` = 6.02 xx 10^23`
No. of molecules of `NH_3` in 4.25g
` = (6.02 xx 10^23)/(17) xx 4.25 = 1.5 xx 10^23`


Discussion

No Comment Found

Related InterviewSolutions