Question

A 0.268g sample of a polypeptide was dissolved in water and made up to a volume...

A 0.268g sample of a polypeptide was dissolved in water and made up to a volume of 160mL at 31.0°C. At this temperature the osmotic pressure of this solution was 4.41mmHg. Calculate the molecular mass of the polypeptide

9.48 g mol-1

7204 g mol-1

729912 g mol-1

1152 g mol-1

960 g mol-1
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Answer #1

Osmotic pressure = CRT, C= concentration = moles/Liter, moles= mass/molar mass, let M= molar mass, Volume = 160ml= 160/1000 L=0.16L

moles= 0.268/M, concentration =0.268/M*0.16 =1.675/M

R=0.0821 L.atm/mole.K, T=31+273= 304 K, Osmotic presure= 4.41 mmHg= 4.41/760 atm =0.0058 atm

hecne 0.0058 = (1.675* 304*0.0821 /M)

M= 1.675*304*0.0821/0.0058 = 7205 gm/mole ( B is correct)

A 0.268g sample of a polypeptide was dissolved in water and made up to a volume of 160mL at 31.0°C. At this temperature the osmotic pressure of this solution was 4.41mmHg. Calculate the molecular mass of the polypeptide

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