Question

45.7 g of magnesium chloride (MgCl2) is dissolved in 2.40 kg of water.
(a) What is the molality (m) of the solution?
(b) How many moles of MgCl2 are contained in 1.76 kg of solvent?

Answer 1

a:- 0.316 molal.

b:- ......

Answer 2

Answer:

a) The solution's molality (m) calculated = $0.2molal$.

b) The number of moles of $MgCl_{2}$ in $1.76 kg$ of solvent = $0.35mol$

Explanation:

Given,

The mass of magnesium chloride ( $MgCl_{2}$ ), solute = $45.7g$

The mass of solvent, i.e., water = $2.40kg$

a) The molality (m) of the solution =?

As we know, the molar mass of $MgCl_{2}$ = $95.2g/mol$

Now,

•  Number of moles = $\frac{Given mass}{Molar mass}$ = $\frac{45.7}{95.2}$ = $0.48mol$

Now,

• Molality = $\frac{Number of moles}{Mass of solvent (kg)}$ = $\frac{0.48}{2.40}$ = $0.2molal$

b)  The number of moles of $MgCl_{2}$ are contained in $1.76 kg$ of solvent =?

As given,

• Mass of solvent = $1.76kg$

As calculated above,

• Molality = $0.2molal$

Therefore,

• The number of moles = mass of solvent × molality = $1.76\times 0.2$ = $0.35mol$