Question

g An aqueous solution of 2.35 M urea is made. The density of the solution is 1.029 g/mL. What is the molality (molal concentration) of urea in the solution. (Molar mass of urea

Answer 1

The molality of the urea solution is 2.35 moles / 0.888 kg = 2.65 mol/kg.

The molality of a solution is calculated by the number of moles of solute divided by the mass of the solvent in kilograms.

In this case, we have 2.35 moles of urea in 1000 mL of solution. The mass of the solution is:

1000 mL * 1.029 g/mL = 1029 g.

The molar mass of urea is approximately 60 g/mol, the mass of urea in the solution is:

2.35 moles * 60 g/mol = 141 g.

Therefore, the mass of the water in the solution is 1029 g - 141 g = 888 g, or 0.888 kg.

So, the molality of the urea solution is:

2.35 moles / 0.888 kg = 2.65 mol/kg.

Note that this calculation assumes that the volume of the solution is not significantly affected by the addition of urea.

Answer 2

1.06 m

Step-by-step explanation:

We have 2.35 moles of pure urea in 1000 ml solution

Mass of the solution = 1000 × 2.35 = 2350 g

Mass of pure urea = 2.35 ×60 = 141 g

Mass of water= 2350 - 141 = 2209 g

Molality = 2.35 / 2209 × 1000 =1.06 m