Question

40 g $MgO$ dissolves in ${{H}_{2}}O$ to form 200 ml of solution, if the density of solution is 1.5 g/ml. Calculate the molality of the solution.

Answer 1

Hint The molality of the solution is one of the major factors in calculating the concentration of the solution. The factors which can be used to calculate the molality of the solution are the number of moles of the solute and the mass of the solvent in kg. With the help of the density of the solution, the mass of the solution can be calculated.

Complete step by step answer:
The concentration of the solution tells the strength of the solution. There are many factors that can be used for calculating the strength of the solution like molarity, molality, normality, mole fraction, ppm, etc. So, the molality of the solution is one of the major factors in calculating the concentration of the solution. There are two factors that are used for the calculation of the molality of the solution, i.e., the number of moles of the solute and the mass of the solvent in kg. When we divide the number of moles of the solute by the mass of the solvent in kg, we get the molality of the solution. The formula is given below:
$Molality=\dfrac{Moles\text{ }of\text{ }the\text{ }solute}{Mass\text{ }of\text{ }the\text{ }solvent\text{ }in\text{ }kg}$
For calculating the moles of the solute, we have to divide the given mass of the solute by the molecular mass of the solute. The formula is given below:
$Moles=\dfrac{Given\text{ }mass}{Molecular\text{ }mass}$
So, the given mass of $MgO$ is 40 g and the molecular mass is 40 g/mol, the number of moles will be:
$Moles=\dfrac{40}{40}=1\text{ mol}$
The volume of the solution is 200 ml, and we know that the density of the solution will be equal to the ratio of mass to the volume.
Mass will be density multiplied by the volume.
$Mass=1.5\text{ x 200 = 300 g}$
300 g is the mass of the solution and 40 g is the mass of the solute, therefore the mass of the solvent will be:
$300-40=260\text{ g}$
260 g = 0.260 Kg
The molality will be:
$Molality=\dfrac{1}{0.260}=3.84\text{ m}$

Therefore, the molality is 3.84 m.

Note: Since there is no volume in the calculation of the molality of the solution, so the molality is temperature independent and does not change even on changing the temperature of the solution.