Question

What is the molality of Ammonia in a solution containing 0.85 g of NH3 in 100 cm3 of a liquid of density 0.85 g/cm3

Answer 1

find the formula:

$n=\frac{m}{M}$

find the mass of solution

$m(s)= 100 cm³ \times0.85 g /cm³ = 85.0 g$

find mass of the water: the solution contains 0.85 g NH3:

$m(w)= 85.0 g (s)- 0.85 g (NH3) = 84.15 g (w)$

find mass of NH3 in 1000 g water:

$m(NH3)=\frac{ 1000 g}{84.15 g}\times 0.85 g (NH3) = 10.1 g (NH3)$

molar mass NH3 = 17.0306 g/mol

find Mol NH3 in 10.1 g:

$n= \frac{10.1g}{ 17.0306 g/mol} = 0.593 mol NH3$