You prepare a solution by dissolving 50.4g sucrose (C12H22O11) in 0.332 kg of water. The final...

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You prepare a solution by dissolving 50.4g sucrose (C12H22O11) in0.332 kg of water. The final volume of the solution is 355 mL.Calculate the concentration of the solution in each unit.
1 mole C12H22O11 = 342.296 g sucrose
(a) Molarity
(b) Molality
(c) Percent by mass
(d) Mole fraction

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4.2 Ratings (511 Votes)

a)
number of moles of Sucrose = mass/molar mass
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 50.4 g / 342.296 g
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.147 mol

Volume = 0.355 L

Molarity = number of moles / Volume
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.147 mol / 0.355 L
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.415 M

b)
Molality = number of moles / mass of solvent in Kg
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.147 mol / 0.332 Kg
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.443 molal

c)
percent by mass = mass of solute *100 / total mass
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 50.4 *100 / (50.4 + 332)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 13.2 %

d)
number of moles of water = mass/ molar mass = 332/18 =18.44 moles

Mole fraction = number of moles of solute / total moles
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.147 / (0.147 + 18.44)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 7.91*10^-3

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