Question

A concentrated solution of sucrose, C12H22O11, contains 233.2 g dissolved in 397.07 g water. The density of the solution is 1.1507 g mL-1 Calculate; weight percent, mol fraction, molality, and molarity of C12H22O11 in the solution

Answer 1

Answer – We are given, mass of sucrose = 233.2 g , mass of water = 397.07 g

Density of solution = 1.1507 g/mL .

Total mass of solution = 233.2 + 397.07 = 630.3 g

Volume of solution = mass / density

                                = 630.3 g / 1.1507 g.ml^-1

                                = 547.7 mL

Moles of sucrose = 233.2 g / 342.29 g.mol^-1

                            = 0.681 moles

Moles of water = 397.07 g / 18.015 g.mol^-1

                          = 22.04 moles

Now weight percent

Weight percent for sucrose = 233.2 g / 630.3 g * 100 %

                                            = 37.0 %

Mole fraction sucrose = 0.681 moles / 22.04 +0.681moles

                                  = 0.03

Molarity of sucrose = 0.681 moles / 0.5477 L

                                  = 1.24 M

Molality of sucrose = 0.681 moles / 0.39707 kg

                                 = 1.752 m