Question

100 mL of an aqueous solution of protein
contains 6.3 g of protein. Calculate the molar mass
of protein if osmotic pressure of the solution at 27°C
is 2.57 x 10-3 bar.​

Answer 1

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Answer 2

Answer:

1 bar = 1 atm

P = 2.57×10^-3 bar

= 2.57 × 10^-3

k = °C + 273

T = 27°C

= 27 + 273

= 300 k

1000 ml = 1 l

100 ml = (1÷1000)×100

100 ml = 0.1 l

V = 0.1 l

R = 0.0821 atm.l/ mol.k

PV = nRT

n = (PV)÷(RT)

n =( 2.57×10^-3 ×0.1)÷(0.0821×300)

n = 0.000257÷24.63

n = 0.0000104344

n = given mass ÷ molar mass

molar mass = given mass ÷ n

= 6.3÷ 0.0000104344

= 603772.1 gm

= 603.7 kg

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