Question

What is the molality of a solution containing 200 mg of urea (molar mass 60g$mo{l^{ - 1}}$) dissolved in 40 g of water?

Answer 1

The molality of the solution is calculated by dividing the number of moles of solute by mass of solvent in kilograms. The number of moles is calculated by dividing mass of the compound by the molecular weight of the compound.

Complete step by step answer:
Given,
Mass of urea is 200 mg.= 0.2 grams
Molar mass of urea is 60 g $mo{l^{ - 1}}$
Mass of water is 40 g. =0.04 kg
Molality is defined as the number of moles of solute dissolved in 1 kg of solvent. The molality is expressed in terms of mol/kg. It is expressed as a molal solution.
The formula to calculate the number of moles is given below.
$n = \dfrac{m}{M}$
n is number of moles.
m is given mass.
M is molecular weight
To calculate the moles of urea, substitute the values mass and molecular weight in the above equation.
$\Rightarrow n = \dfrac{{0.2g}}{{60g/mol}}$
$\Rightarrow n = 0.0033$mol.
The formula to calculate molality is shown below.
$m = \dfrac{n}{{M\;in\;kg}}$
Where,
m is the molality of the solution.
n is the number of moles.
M is the mass of solvent in Kg.
To calculate the molality of the solution, substitute the values of moles and mass in kg in the above equation.
$\Rightarrow m = \dfrac{{0.0033mol}}{{0.04kg}}$
$\Rightarrow m = 0.0825mol/kg$
Therefore, the molality of solution containing 200 mg of urea (molar mass 60g$mo{l^{ - 1}}$) dissolved in 40 g of water is 0.085 m.
Therefore, the correct option is D.

Note:
Make sure to convert the mass in mg into grams 1 milligram = 0.001 gram and convert the mass of urea into kg as molality is expressed in mol/kg. 1g = $\dfrac{1}{{1000}}kg$.