Question

A solution containing 28.60 mg of an unknown protein per 29.0 mL solution was found to have an osmotic pressure of 3.80 torr at 15 ∘C. What is the molar mass of the protein?

Please show details on how to get to the answer:

Mmol =

4660

g/mol

Answer 1

Solution -

Formula to calculate the osmotic pressure is as follows

=MRT

= osmotic pressure

M= molarity

R= 0.08206 L atm per K mol

T= kelvin temperature

convert pressure from torr to atm

3.80 torr * 1 atm / 760 torr = 0.005 atm

T 15 C +273 = 288 K

now lets put the values in the formula

0.005 atm = M* 0.08206 L atm per mol K * 288 K

M = 0.005 atm / 0.08206 L atm per mol K * 288 K

M = 0.000212

Therefore molarity = 0.000212

now using hte molarity lets calculate the moles of the protein

moles = molarity * volume in liter

          = 0.000212 mol per L * 0.029 L

         = 6.1354*10^-6 mol

now lets calculate the molar mass of the protein

molar mass = mass in gram / moles

                   = (28.60 mg * 1 g / 1000 mg) / 6.1354*10^-6 mol

                  = 4661 g/mol

Therefore molar mass of the protein is 4661 g per mol or we can round it to 4660 g per mol