If the osmotic pressure is 1.25 torr, what is the molecular mass of the protein?

Question

24 mg of a certain protein was dissolved in water to make a solution of 1.00 mL at 22C

Answer 1

24 mg=0.024 g
1.00 mL=0.001 L
1.25 torr=0.00164 atm
22 °C=295 K
R=0.08206
MM= (0.024) (0.08206) (295) / (0.00164) (0.001)=
354268 g /mol

Answer 2

I guess you can use my question as a ref.


22 mg of a certain protein was dissolved in water to make a solution of 4.50 mL at 26C. If the osmotic pressure is 1.25 torr, what is the molecular mass of the protein?

To enter an answer in scientific notation use an e




6.022 x 1023 = 6.022e23



1.340 x 10-8 = 1.340e-8


Enter a numeric answer only, do not enter units.

Correct, computer gets: 72931.7185422222

This is how I solved it:

(0.022)*(0.082)*(273+26)/((1.25/760)*0.0045)=72931.7185422222

Answer 3

πV=nRT
π=osmotic pressure= 1.25 torr=1.25/760 atm
V= vol of soln.= 1ml = 1x 10^-3 lt
n= no of moles of solute.
R= universal gas const= 0.0821 lt atm mole^-1 K^-1
T= 22 deg C = 299 K
πV=nRT
or,πV=WRT/M...........(2)
W = Wt of solute = 24 mg = 24x 10^-3 g
M = molecular mass of protein

M= WRT/πV
= 24x 10^-3x0.0821x299/(1.25/760)x(1x 10^-3)
=358202.95(ans)