Question

An aqueous solution of 10.03 g of catalase, an enzyme found in the liver, has a volume of 1.05 L at 27°C. The solution's osmotic pressure at 27°C is found to be 0.745 torr. Calculate the molar mass of catalase._________ g/mol

Answer 1

Answer : The molar mass of catalase is, $2.40\times 10^5g/mol$

Explanation :

Formula used :

$\pi =CRT\\\\\pi=\frac{w}{M\times V}RT$

where,

$\pi$ = osmotic pressure  = 0.745 torr = 0.000980 atm   (1 atm = 760 torr)

C = concentration

R = solution constant  = 0.0821 L.atm/mol.K

T = temperature  = $27^oC=273+27=300K$

w = mass of catalase = 10.03 g

M = molar mass of catalase = ?

V = volume of solution  = 1.05 L

Now put all the given values in the above formula, we get:

$0.000980atm=\frac{10.03g}{M\times 1.05L}\times (0.0821L.atm/mole.K)\times (300K)$

$M=2.40\times 10^5g/mol$

Therefore, the molar mass of catalase is, $2.40\times 10^5g/mol$