When 1.32 g of nonpolar solute was dissolved In 50.0 g of phenol, the latter\'s

Question

When 1.32 g of nonpolar solute was dissolved In 50.0 g of phenol, the latter's freezing point WAS Lowered by 1.454 degree C calculate the motor man of the solute. K(for phenol is 7.27 degree C/m Intravenous medications an often administered in 5% percent glucose, C_6H_12O_6(aq). by mass. What is the osmotic pressure of such solutions at 37 degree C? Assume that the density of the solution is 1.0 g/mL. A 262 0 mL sample of a sugar solution containing I 22 s of the sugar has an osmotic pressure of 30.3 mm Kg at 35 degree C. What is the molar mass of the sugar?

Explanation / Answer

6) The depression in freezing point = Tf X molality

Molality = Moles / Mass of solvent = Mass of solute / Molecular weight of solute X mass of solvent in Kg

so

1.454 = 7.27 X 1.32 / Mol wt X 0.05 Kg

Mol wt = 132 g / mole

7) Osmotic pressure = Concentration X RT

R = 0.0821

T = 310 K

concentration = Moles of solute / Volume of solution

Let the mass of solution is 100 grams so mass of solute = 5g

So moles = 5 / 180 = 0.0278

Density is 1g/mL

So volume of solution = 10mL = 0.1 L

concentration = 0.0278 / 0.1 = 0.278

Osmotic pressure = Concentration X R X T = 0.278 X 0.0821 X 310 = 7.07 atm

8) Again

Osmotic pressure = Concentration X RT = Moles of solute X R X T / Volume of solution

Moles of solute = Osmotic pressure X volume of solution / RT

osmotic pressure = 30.3 mmHg = 0.0398 atm

Volume = 0.262 L

T = 35 + 273 K

So moles = 0.0398 X 0.262 / 0.0821 X 308 = 4.12 X 10^-4 moles

Molecular weight = Mass / Moles = 1.22 / 4.12 X 10^-4 = 296 grams/ Mole

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