Question

33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b. What is the mole fraction of fructose in this solution? c. What is the solution's average molar mass? d. What is the specific molar volume of the solution?

Answer 1

Step-by-step explanation:

Mass of fructose = 33.56 g

Mass of water = 18.88 g

Total mass of the solution = Mass of fructose + Mass of water = M

M = 33.56 g + 18.88 g =52.44 g

Volume of the solution = V = 40.00 mL

Density =
`(Mass)/(Volume)`

a) Density of the solution:

`(M)/(V)=(52.44 g)/(40.00 mL)=1.311 g/mL`

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose =
`n_1=( 33.56 g)/(180.16 g/mol)=0.1863 mol`

Molar mass of water = 18.02 g/mol

Moles of water=
`n_2=( 18.88 g)/(18.02 g/mol)=1.0477 mol`

Mole fraction of fructose in this solution:
`\chi_1`

`\chi_1=(n_1)/(n_1+n_2)=(0.1863 mol)/(0.1863 mol+1.0477 mol)`

`\chi_1=0.1510`

Mole fraction of water =
`\chi_2=1-\chi_1=0.8490`

c) Average molar mass of of the solution:

=
`\chi_1* 180.16 g/mol+\chi_2* 18.02 g/mol`

`=0.1510* 180.16 g/mol+0.8490* 18.02 g/mol=42.50 g/mol`

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

`\frac{\text{Average molar mass}}{\text{Density of the mass}}`

`=(42.50 g/mol)/(1.311 g/mL)=32.42 mL/mol`