A solution of protein (extracted from crab) was prepared by dissolving 0.75g in 125ml of a solution. At 4°C an osmotic pressure rise of 2.6 mm of the solution was observed. Then molecular mass of protein is (assume density of solution is 1gm/ml) ?

1 Answer
May 12, 2018

f.wt= 39,878(g)/("mole")~~4bar0,000(g)/("mole")

Explanation:

Pi=MRT = (("mass"(g))/(f.wt.))V^-1*R*T
=> f.wt.="mass"(g)*Pi^-1*V^-1*R*T
Pi => Osmotic Pressure (Atm) = (2.6/760)Atm = 0.0034Atm
R => Gas Constant = 0.08206(L*Atm)/("mole"*K)
T => Temperature (Kelvin) = (4+ 273)K = 277K
"mass"(g)=0.75g
V=Volume(L)=0.125L

f.wt. = (0.75g)/((0.00342cancel(Atm)*0.125cancel(L)))*0.08206(cancel(L)*cancel(Atm))/("mole"*cancel(K))*277cancel(K)

= 39,878(g)/("mole")~~4bar0,000(g)/("mole")