Question

A solution is prepared by dissolving 50.8 g sucrose(C12H22O11) in 0.387 kg of water. The final volume of the solution is 355 mL.

For this solution, calculate molarity.

For this solution, calculate molality.

For this solution, calculate percent by mass.

For this solution, calculate mole fraction.

Answer 1

Molarity= wt of solute/mol.wt X 1000/V of sol. in litre

sucrose wt = 50.8 g

sucrose mol wt = 342.29 g/mol

solution voleme (V) = 355 ml.

M = 50.8/342.29 x 1000/355

Molarity = 0.418 M.

molality:

molality (m) = wt/ molwt X 1/wt of solvent in kg.

solvent weight = 0.87 kg

molality = 50.8/342.29 X 1/0.87

molality = 0.170m.

percent by mass:

percent by mass = solute weight x 100/ weight of solution solute is the sucrose

solution weight = solute weight + solvent weight

solution weight = 50.8 + (0.387 x 1000 g)

solution weight = 437.8 g

percent by mass = solute weight x 100/ weight of solution

percentage by mass = 50.8 x100/437.8

percentage mass = 11.60

mole fraction:

no of moles of solute = n   

no of moles of solvent =N

mole fraction for solute = n/n+N

sucrose :

no of moles (n) = wt/ molecular weight

n = 50.8/342.29

   n = 0.1484

Water:

no of moles (N) = wt/ molecular weight

N = 0.387 x 1000g/18

N = 21.5

molefraction of sucrose:

= n/n+N

= 0.1484/0.1484+21.5

   = 0.1484/21.64

   = 0.00685

molefraction of water:

= N/N+n

= 21.5/21.5+0.1484

=21.5/21.6484

=0.993