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

Step-by-step explanation:

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

`Molality=(n* 1000)/(W_s)`

where,

n= moles of solute

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

`W_s` = weight of solvent in g

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

Putting in the values we get:

`Molality=(0.5* 1000)/(500)=1mole/kg`

Thus molality of solution will be 1 molal.