Referring to the van\'t Hoff factors in the table below, calculate the mass of s

Question

Referring to the van't Hoff factors in the table below, calculate the mass of solute required to make each aqueous solution. Table. Van't Hoff Factors at 0.05 m Concentration in Aqueous Solution Solute i Expected i Measured Nonelectrolyte 1 1 NaCl 2 1.9 MgSO4 2 1.3 MgCl2 3 2.7 K2SO4 3 2.6 FeCl3 4 3.4 A) a sodium chloride solution containing 167 g of water that has a melting point of -1.9 C. B)274 mL of a magnesium sulfate solution that has an osmotic pressure of 3.88 atm at 298 K C)an iron(III) chloride solution containing 229 g of water that has a boiling point of 107 C

Explanation / Answer

1) Depression in freezing point = i x Kf x m

for water Kf = 1.86

0 + 1.9 = 1.9 x 1.86 x m

m = 0.537

Knowing that

molality = moles of solute / mass of solvent (Kg)

0.537 = moles of NaCl / 0.167

moles of Nacl = 0.0896

moles = mass / molar mass

mass of NaCl = 0.0896 x molar mass

mass of NaCl = 0.0896 x 58.44

mass of NaCl = 5.23

so 5.23 g of NaCl is required

2)

osmotic pressure

pi = i x C x R x T

so

3.88 atm = 1.3 x C x 0.0821 atmL/molK x 229 K

C = 0.1215 M

molarity = moles / volume (L)

so

0.1215 M = moles of MgS04 / 0.274

moles of MgS04 = 0.0332 mol

mass of MgS04 = 0.0332 mol x molar mass

= 0.0332 mol x 120.366

= 3.99 g

so 3.99 g of MgS04 is required

3) Elevation in boiling point = i x kb x m

for solvent water Kb = 0.512

so

(107 - 100)K = 3.4 x 0.512 x m

m = 4.020 M

molality = moles of solute / mass of solvent (Kg)

4.020M = moles of feCl3 / 0.229

moles of FeCl3 = 0.98 mol

moles = mass / molar mass

mass of FeCl3 = 0.98 x molar mass

mass of FeCl3 = 0.98 x 162.2

mass of FeCl3 = 158.95

so 158.95 g of FeCl3 is required

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