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Nutka1998 [239]

3 years ago

7

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

1 answer:

-Dominant- [34]3 years ago

7 0

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, $2.40\times 10^5g/mol$

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For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The

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Answer : The rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

Explanation :

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b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

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Putting values in equation 1, we get:

$6.0\times 10^{-4}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-2}M^{-2}s^{-1}$

Thus, the value of the rate constant 'k' for this reaction is $7.5\times 10^{-2}M^{-2}s^{-1}$

Now we have to calculate the rate for trial 5 that starts with 0.90 M of reagent A, 0.60 M of reagents B and 0.70 M of reagent C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-2})\times (0.90)^2(0.60)^0(0.70)^1$

$\text{Rate}=4.25\times 10^{-2}Ms^{-1}$

Therefore, the rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

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