Chemistry, asked by gaurav588677, 10 months ago


Osmotic pressure of a solution contaning 7 gram ofa
protein per 100 cm cube of solution is 3.3x10 minus 2bar at
37°C. Calcualte the molar mass of protein.


LuckyYadav2578: dear please see the answer again ... i m really very sorry... by mistakenly i have putted the value of R = 0.821 which was wrong but i have corrected it ....

Answers

Answered by LuckyYadav2578
5

P ( osmotic pressure ) = 3.3 × 10^(-2)

V ( volume ) = 100 cm^3

w ( weight ) = 7 gram

m ( molar mass ) = ?

R = 0.821

T ( temperature ) = 37 degree Celsius = 310 Kelvin

PV = (w/m) RT

[3.3 × 10 ^(-2)] × (100/ 1000) = (7/m) × 0.0821 × 310

3.3 × 10 ^(-3) = 178.157/m

m = 178157/3.3

m = 53986.9697

m = 5.4 × 10^4 g mol^(-1)

Attachments:
Answered by sehangshu22
0

Explanation:

P ( osmotic pressure ) = 3.3 × 10^(-2)

V ( volume ) = 100 cm^3

w ( weight ) = 7 gram

m ( molar mass ) = ?

R = 0.821

T ( temperature ) = 37 degree Celsius = 310 Kelvin

PV = (w/m) RT

[3.3 × 10 ^(-2)] × (100/ 1000) = (7/m) × 0.821 × 310

3.3 × 10 ^(-3) = 1781.57/m

m = 1781570/3.3

m = 539869.697

m = 5.4 × 10^5 g mol^(-1)

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