Question

A student dissolves 12.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.01/gmL . The student notices that the volume of the solvent does not change when the sucrose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.

Answer 1

Answer :  The molarity and molality of the student's solution is, 0.12 mol/L and 0.12 mol/kg respectively.

Explanation : Given,

Density of solvent = 1.01 g/mL

Molar mass of sucrose = 342.3 g/mole

Mass of sucrose = 12 g

Volume of solvent = 300 mL

First we have to calculate the mass of solvent.

$\text{Mass of solvent}=\text{Density of solvent}\times \text{Volume of solvent}=1.01g/mL\times 300mL=303g$

Now we have to calculate the molarity of solution.

$\text{Molarity}=\frac{\text{Mass of sucrose}\times 1000}{\text{Molar mass of sucrose}\times \text{Volume of solution (in mL)}}$

$\text{Molarity}=\frac{12g\times 1000}{342.3g/mole\times 300mL}=0.12mole/L$

Now we have to calculate the molality.

$\text{Molality}=\frac{\text{Mass of sucrose}\times 1000}{\text{Molar mass of sucrose}\times \text{Mass of water (in g)}}$

$\text{Molality}=\frac{12g\times 1000}{342.3g/mole\times 303g}=0.12mole/kg$

Therefore, the molarity and molality of the student's solution is, 0.12 mol/L and 0.12 mol/kg respectively.