Given:
Mass of solute (NH1) in the solution = 35.0 g
Mass of solvent (water) in the solution = 75.0 g
Density of solution is 0.982 g/ml
The molar mass of ammonia (NH3) is 17.03 g/mol
The number of moles of solute (NH3) is calculated using its molar mass
[tex]\frac{35.0 \: g \:NH_{3}(?) }{17.3 \: g \: mol \: } = 2.055 \: mol \: NH_{3}[/tex]
The mass of solvent (water) =
[tex]75.0g \: h_{2}o \: \times \: \frac{1kg}{ {10}^{3} } kg[/tex]
[tex]75.0g \: h_{2}o[/tex]
