Question

200 ml of aqueous solution of protein contain 1.26 of protein the osmotic pressure of such solution at 300 k is found to be 2.57×10^3. Calculate the molecular mass of protein....... please help urgent.....no spam......give satisfying answer...... thank u

Answer 1

hey
here is ur answer

isi se kaam chalalo likhne ka time nhi
I hope this will help
#Prem✴✌✴✌

Answer 2

Answer:  The molecular mass of protein is $60377.04g/mol$

Explanation:

$\pi =CRT$ if osmotic pressures are equal at the same temperature, concentrations must also be equal.

$\pi$ = osmotic pressure  =$2.57\times 10^{-3}atm$

C= concentration

R= solution constant  = 0.0821 Latm/Kmol

T= temperature  = 300 K

For protein: 1.26 g of urea is dissolved in 200 ml of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}$

$C_{protein}=\frac{1.26\times 1000}{200\times M_{protein}}$

$2.57\times 10^{-3}=\frac{1.26\times 1000}{M_{protein}\times 200}\times 0.0821\times 300K$

$M_{protein}=60377.04g/mol$