Question

A solution is prepared by dissolving 15.0 g of NH3 in 250 g of water. The density of the resulting solution is 0.974 g/mL. The mole fraction of NH3 in the solution is _____.A) 16.8B) 0.940C) 0.0597D) 0.922E) 0.0640

Answer 1

Mole fraction of
`NH_(3)` in solution is 0.0597.

Step-by-step explanation:

Molar mass of
`NH_(3)` = 17.031 g/mol

Molar mass of
`H_(2)O` = 18.015 g/mol

No. of moles = (mass)/(molar mass)

So, 15.0 g of
`NH_(3)` =
`(15.0)/(17.031)` moles of
`NH_(3)` = 0.8807 moles of
`NH_(3)`

250 g of
`H_(2)O` =
`(250)/(18.015)` moles of
`H_(2)O` = 13.88 moles of
`H_(2)O`

So, Mole fraction of
`NH_(3)` in the solution = (no.of moles of
`NH_(3)` )/(total no. of moles) =
`(0.8807)/(0.8807+13.88)` = 0.0597

Hence, mole fraction of
`NH_(3)` in solution is 0.0597.