Question

A student dissolves 15.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.13/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. Be sure each of your answer entries has the correct number of significant digits

Answer 1

Answer: The molarity and molality of sucrose solution is 0.146 M and 0.129 m respectively

Explanation:

• Calculating the molarity of solution:

To calculate the molarity of solution, we use the equation:

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

Given mass of sucrose = 15 g

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 300 L

Putting values in above equation, we get:

$\text{Molarity of sucrose solution}=\frac{15\times 1000}{342.3\times 300}\\\\\text{Molarity of sucrose solution}=0.146M$

Hence, the molarity of sucrose solution is 0.146 M

• Calculating the molality of solution:

To calculate the mass of solvent, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solvent = 1.13 g/mL

Volume of solvent = 300 mL

Putting values in above equation, we get:

$1.13g/mL=\frac{\text{Mass of solvent}}{300mL}\\\\\text{Mass of solvent}=(1.13g/mL\times 300mL)=339g$

To calculate the molality of solution, we use the equation:

$\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute (sucrose) = 15 g

$M_{solute}$ = Molar mass of solute (sucrose) = 342.3 g/mol

$W_{solvent}$ = Mass of solvent = 339 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{15\times 1000}{342.3\times 339}\\\\\text{Molality of solution}=0.129m$

Hence, the molality of sucrose solution is 0.129 m