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
Doc048
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")#

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