Question

A solution containing 292 g of Mg(NO3)2 per liter has a density of 1.108 g/mL. The molality of the solution is:

A) 2.00 m

B) 1.77 m

C) 6.39 m

D) 2.41 m

E) none of these

Answer 1

Answer: D) 2.41 m

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molality=\frac{n}{W_s}$

where,

n = moles of solute

 $W_s$ = weight of solvent in kg

moles of solute =$\frac{\text {given mass}}{\text {molar mass}}=\frac{292g}{148g/mol}=1.97moles$

volume of solution = 1L = 1000 ml      (1L=1000ml)

Mass of solution=${\text {Density of solution}}\times {\text {Volume of solution}}=1.108g/ml\times 1000ml=1108g$

mass of solute = 292 g

mass of solvent = mass of solution - mass of solute = (1108- 292) g = 816g = 0.816 kg

Now put all the given values in the formula of molality, we get

$Molality=\frac{1.97moles}{0.816kg}=2.41mole/kg$

Therefore, the molality of solution will be 2.41 mole/kg