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goldfiish [28.3K]
3 years ago
8

β‑Galactosidase (β‑gal) is a hydrolase enzyme that catalyzes the hydrolysis of β‑galactosides into monosaccharides. A 0.387 g sa

mple of β‑galactosidase is dissolved in water to make 0.137 L of solution, and the osmotic pressure of the solution at 25 ∘ C is found to be 0.602 mbar. Calculate the molecular mass of β‑galactosidase.molecular mass: ____ g/mol.
Chemistry
1 answer:
gtnhenbr [62]3 years ago
3 0

Answer:

The molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=icRT

where,

\pi = osmotic pressure of the solution = 0.602 mbar = 0.000602 bar

0.000602 bar = 0.000594 atm

(1 atm = 1.01325 bar)

i = Van't hoff factor = 1 (for non-electrolytes)

c = concentration of solute = ?

R = Gas constant = 0.0820\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273.15 +25]=298.15 K

Putting values in above equation, we get:

0.000594 atm=1\times c\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298.15 K\\\\c=2.4278\times 10^{-5} mol/L

The concentration of solute is 2.4278\times 10^{-5} mol/L

Volume of the solution = V =0.137 L

Moles of β‑Galactosidase = n

C=\frac{n}{V(L)}

n=2.4278\times 10^{-5} mol/L\times 0.137 L

n=3.3261\times 10^{-6} mol

To calculate the molecular mass of solute, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of β‑Galactosidase = 3.3261\times 10^{-6} mol

Given mass of β‑Galactosidase= 0.387 g

Putting values in above equation, we get:

3.3261\times 10^{-6} mol =\frac{0.387 g}{\text{Molar mass of solute}}\\\\\text{Molar mass of solute}=116,352.97 g/mol

Hence, the molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

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What is the rate law for the reaction 2A + 2B + 2C --&gt; products
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R = 47.19 [A]*([B]^2)*[C]

Explanation:

The rate law for the reaction 2A + 2B + 2C --> products

Is being sought.

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where m, n, and p are the reaction orders with respect to (w.r.t.) components A, B and C respectively. This could be reduced to

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Where ka=(k[B]^n)*([C]^p);    

R = kb ([B]^n)                    (3)

Where kb=(k[A]^m)*([C]^p); and  

R = kc ([C]^p)                     (4)

Where kc=(k[A]^m)*([B]^n).

Equations (2), (3) and (4) are obtained for cases when the concentrations of two components are kept constant, while only one component’s concentration is varied. We can determine the reaction wrt each component by employing these equations.  

The readability is very much enhanced when the given data is presented in the following manner:

Initial [A]  0.273   0.819   0.273   0.273

Initial [B]  0.763   0.763   1.526   0.763

Initial [C]  0.400   0.400    0.400   0.800

Rate           3.0       9.0       12.0       6.0

Run#  1  2 3  4

Additional row is added to indicate the run # for each experiment for easy reference.

First, we use the initial rate method to evaluate the reaction order w.r.t. each component [A], [B] and [C] based on the equations (2), (3) and (4) above.

Let us start with the order wrt [A]. From the given data, for experimental runs 1 and 2, the concentrations of reactants B and C were kept constant.

Increasing [A] from 0.273 to 0.819 lead to the change of R from 3.0 to 9.0, hence we can apply the relation based on equation (2) between the final rate R2, the initial rate R1 and the final concentration [A2] and the initial concentration [A1] as follows:

R2/R1=ka[A2]^m/ka[A1]^m=([A2]/[A1])^m

9.0/3.0 = (0.819/0.273)^m

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Similarly, applying experimental runs 1 and 3 could be applied for the determination of n, by employing equation (3):  

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And finally for the determination of p we have using runs 4 and 1:

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6/3= (0.8/0.4)^p

2= 2^p , -> p = 1

Therefore, plugging in the values of m, n and p into equation (1), the rate law for the reaction will be:

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The value of the rate constant k could be estimated by making it the subject of the formula, and inserting the given values, say in run 1:

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47.19

Finally, the rate law is

R = 47.19 [A]*([B]^2)*[C]

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