Camphor (C_10H_16O) freezes at 179.8 degree C, and it has a particularly large f

Question

Camphor (C_10H_16O) freezes at 179.8 degree C, and it has a particularly large freezing-point depression constant (K_fp) of 40.0 degree C/m. When 0.186 g of an no electrolyte organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 degree C. Calculate the molar mass of the solute. A sample of 350.0 mg of a purified protein was dissolved in 10.00 mL of solution. The osmotic pressure was found to be 10.0 mm Hg at 25.0 degree C. Calculate the molar mass of the protein.

Explanation / Answer

Answer :

3.

Calculation molar mass of organic compound :

We are give freezing point of solution and solvent. We know delta Tf = 179.8-176.7 = 3.1 0C

We know depression in freezing point (Delta Tf) = m kf

Here kf is freezing point constant of solvent (camphor here ) . m is molality of the solute.

Molality = number of moles of solute / volume of solvent in kg.

Number of moles = Mass in g / molar mass

Lets use these formula and given value to calculate molar mass of organic compound.

3.1 0C = [(0.186 g /Molar mass ) / 0.02201 kg ) ] x 40.0 deg C kg /m

Lets find molar mass

Molar mass = 109.04 g /mol

4 :

We use following formula to calculate molar mass of protein.

p= C RT

Here C is mol /L , R is gas constant and T is temperature in K

C = ( 0.10/760 ) / ( 0.08206 x 298.15 )

C = 0.01358

0.01358 = ( 0.350 / MM )/0.010 L

MM of potein = 650.6 g /mol

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