Question

A sucrose solution is prepared to a final concentration of 0.190 M . Convert this value into terms of g/L, molality, and mass % (molecular weight, MWsucrose = 342.296 g/mol ; density, ρsol′n = 1.02 g/mL ; mass of water, mwat = 955.0 g ). Note that the mass of solute is included in the density of the solution.

Answer 1

The strength of the sucrose solution is 65.03624 g/L.

Molality of the solution is 0.1989 mol/kg.

Mass percentage of the solution is 6.37%

Step-by-step explanation:

Strength of the sucrose solution

Molarity of the sucrose solution = 0.190 M = 0.190 mol/L

Molar mass of sucrose = 342.296 g/mol

Strength of the solution = Molarity of the solution × Molar mass of the solute

Strength of the sucrose solution =

=
`0.190 mol/L* 342.296 g/mol=65.03624 g/L`

Molality of the solution:

`Molality=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}`

Moles of sucrose in 1L solution = 0.190 moles

Mass of water = 955.0 g = 0.9550 kg

Molality of the solution is:

`Molality=(0.190 mol)/(0.9550 kg)=0.1989 mol/kg`

Mass percentage of the solution:

Density iof the solution =
`\rho _(sol'n)=1.02 g/mL`

Mass of solution = Mass of solute + Mass of water

Volume of solution = 1L = 1,000 mL

`\rho _(sol'n)=1.02 g/mL=\frac{\text{mass of solution}}{\text{Volume of solution}}`

Mass of solution = 1.02 g/mL × 1,000 mL = 1,020 g

Mass of solute + Mass of water = 1,020 g

Mass of solute = 1,020 g - 955.0 g = 65.0 g

Mass percentage:

`\%(w/w)=\frac{\text{mass of solute}}{\text{mass of solution}}* 100`

`\%(w/w)=(65.0 g)/(1,020 g)* 100=6.37\%`

Mass percentage of the solution is 6.37%

Answer 2

Molarity is moles / L solution. Therefore final concentration is 0.190 moles / L.

A. Converting to g / L

(0.190 moles / L) * (342.296 g/mol) = 65.04 g / L

B. Converting to molality = moles / kg water

(0.190 moles / L) (1 L / 0.955 kg water) = 0.20 molal

C. Converting to mass%

mass% = (0.190 moles / L) * (342.296 g/mol) * (1 mL / 1.02 g) * (1 L / 1000 mL) * 100%

mass% = 6.38%