Question

1.) A 0.700 g sample of a nonvolatile yellow crystalline solid is dissolved in 16.3 g of benzene, producing a solution that freezes at 5.10oC. Find the molar mass of the yellow solid in g/mol.

2.) What is the osmotic pressure in atm of a 0.130 M solution of NaCl at 0^oC?

3.) Calculate the freezing point for a 1.55 m solution of CCl₄ in benzene.

Answer 1

Mass of solute = 0.7 g

Mass of solvent, benzene = 16.3 g = 0.0163 kg

Freezing point of pure benzene = 5.5 ^oC

Freezing point of solution = 5.1 ^oC

depression in freezing point = $\Delta T_{f}$ = 5.5 - 5.1 = 0.40 ^oC

K_f of benzene = 4.90 ^oC/m

Formula

$\Delta T_{f}=K_{f}\times molality$

Solving for molality

Plug in the values

0.40 = 4.90 x m

molality = 0.082 m

But molality = moles/ weight of solvent

0.082 = moles/0.0163

multiply both sides with 0.0163

moles of solute = 0.082*0.0163 = 1.33 x 10^-3 mol

Finding molar mass of solute

Molar mass of solute = weight /moles

= 0.7 g/1.33 x 10^-3 mol

= 526.3 g/mol

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2)

Concentration of NaCl solution = 0.130 M

Number of ions in NaCl = i = 2 (Na+ and Cl- ions)

R = 0.0821 L.atm/mol.K

Temperature = T = 0 ^oC = 273 K

Formula for osmotic pressure

$\pi =i \times c\times R\times T$

$\pi =2\times0.130\times 0.0821\times 273$

$\pi =5.83 atm$

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3)

Given, molality of solution = 1.55 m

Kf = 4.90 ^oC/m

Formula for depression in freezing point =

$\Delta T_{f}=K_{f}\times molality$

$\Delta T_{f}=4.90\times 1.55$

$\Delta T_{f}=7.60 ^{\circ}C$

Freezing point of pure benzene = 5.5 ^oC

Freezing point of CCl₄ -benzene solution = 5.50 - 7.60 = -2.1 ^oC