Question

Given that the density of water is 0.975 g/mL and that 171 g of sucrose (molar mass: 342.30 g/mol) is dissolved in 512.85 mL of water at 80°C, what is the molality of this solution?

Answer 1

1.00 m is the molality of this solution

Answer 2

Answer: 1 molal

Explanation:

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

$Molality=\frac{n\times 1000}{W_s}$

where,

n= moles of solute

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{171g}{342.30g/mol}=0.5moles$

 $W_s$ = weight of solvent in g

$\text {Mass of solvent}={\text {density of solvent}\times {\text {Volume of solvent}=0.975g/ml\times 512.85ml=500g$

Putting in the values we get:

$Molality=\frac{0.5\times 1000}{500}=1mole/kg$

Thus molality of solution will be 1 molal.