Question

A)

A solution is prepared by dissolving 50.4 g sucrose (C12H22O11) in
0.332 kg of water. The final volume of the solution is 355 mL. For
this solution, calculate the molarity. **Express the molarity
in units of moles per liter to three significant
figures.**

**B)**

Calculate the molality. **Express the molality in units of
moles per kilogram of solvent to three significant
figures.**

**C)**

Calculate the percent by mass.**Express the percent by
mass to three significant figures.**

**D)**

Calculate the mole fraction. **Express the mole fraction
to three significant figures.**

**E)**

Calculate the mole percent. **Express the mole percent to
three significant figures.**

Answer #1

A) Molarity = number of moles per litre of solution

Molecular weight of sucrose is 342 g

= (50.4/342)/(355/1000)

= 0.415 M

B) Molality = (50.4/342)/0.332

= 0.444

C) Mass percent = (50.4/(332 + 50.4))*100 = 13.2 %

D) Number of moles of sucrose = (50.4/342) = 0.1474

Number of moles of waater = 332/18 = 18.4444

Thus mole fraction of sucrose = 0.1474/( 0.1474 + 18.4444 )

= 0.008

Mole fraction of water = 1 - 0.008 = 0.992

E) Mole percent of sucrose = mole fraction * 100 = 0.8 %

Mole percent of water = mole fraction * 100 = 99.2 %

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