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 ....
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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)
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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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