34.2g of sugar was dissolved in water to prapare124.2g of sugar syrup. calculate milality and mole fraction of sugar in the syrup
Answers
1.12 mol / kg ^-1 (approx.) and 0.01960784313
Given:
Mass of solute(sugar) = 34.2 grams
Mass of solution(sugar syrup) = 124.2 grams
To Find:
Molality of the sugar in the syrup
Mole fraction of sugar in the syrup
Solving:
Formula to calculate the mass of solution:
Mass of solution = Mass of solvent + Mass of solute
Here we are going to find the mass of solvent:
Mass of Solvent = Mass of solution - Mass of solvent
= 124.2 g - 34.2 g
= 90 grams
Converting into kilograms:
= 90 / 1000
= 0.09 kg
Calculating the molar mass of Sugar:
(C12H22O11)
= 12 g x 12 + 1 g x 22 + 16 g x 11
= 342 g mol^-1
No of moles of solute
(Sugar)
= Given Mass / Molecular Mass
= 34.2 g / 342 g mol^-1
= 0.1 mole
Calculating Molality of Sugar:
Molality of solution = Number of moles of solute / Mass of solvent in kg
Substituting the values in the formula we get:
= 0.1 mol / 0.09 kg
= 1.12 mol / kg ^-1 (approx.)
Calculating the molar mass of Water:
(H2O)
= 1 g x 2 + 16 g x 1
= 2 g + 16 g
= 18 g mol^-1
Number of moles in water:
= Given Mass / Molecular Mass
= 90 g / 18 g mol^-1
= 5 mol
Calculating total number of moles:
= nA + nB
= 0.1 + 5
= 5.1 mol
Mole fraction of sugar (solute):
= xB = nB/(nA + nB)
= 0.1 / 5.1
= 0.01960784313 (approx.)
Therefore, the molality of sugar is 1.12 mol / kg ^-1 (approx.) and mole fraction of sugar in the syrup is 0.01960784313.