Question

 

A) Camphor has a melting point of 178.4℃ and a K_f of 37.7℃/m. If 0.550 g of an unknown nonelectrolyte lowered the freezing point of 35.0 g of camphor by 2.00℃, what was the molar mass of the unknown?

B) When 1.00 g of hemoglobin is dissolved in enough water to form 100mL of solution, the osmotic pressure at 20℃ is 3.62x10^-3 atm. What is the molar mass of the protein, hemoglobin?

Answer 1

A)

we have below equation to be used:

delta Tf = Kf*mb

2.0 = 37.7 *mb

mb= 0.0531 molal

mass of solvent = 35.0 g

= 3.5*10^-2 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

number of mol,

n = Molality * mass of solvent in Kg

= (0.0531 mol/Kg)*(0.035 Kg)

= 1.857*10^-3 mol

mass of solute = 0.550 g

we have below equation to be used:

number of mol = mass / molar mass

1.857*10^-3 mol = (0.55 g)/molar mass

molar mass = 296 g/mol

Answer: 296 g/mol

B)

P = 0.00362atm

T= 20.0 oC

= (20.0+273) K

= 293 K

we have below equation to be used:

P = C*R*T

0.0036 = C*0.0821*293.0

C =0.0002 M

volume , V = 100 mL

= 0.1 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.0002*0.1

= 1.505*10^-5 mol

mass of solute = 1.00 g

we have below equation to be used:

number of mol = mass / molar mass

1.505*10^-5 mol = (1.0 g)/molar mass

molar mass = 6.64*10^4 g/mol

Answer: 6.64*10^4 g/mol