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Chapter 13: Problem 81

Lysozyme is an enzyme that breaks bacterial cell walls. A solution containing \(0.150 \mathrm{~g}\) of this enzyme in \(210 \mathrm{~mL}\) of solution has an osmotic pressure of \(0.953\) torr at \(25^{\circ} \mathrm{C}\). What is the molar mass of lysozyme?

Short Answer

Expert verified
The molar mass of lysozyme is approximately \(4670 \thinspace g \thinspace mol^{-1}\).

Step by step solution

01

Convert the given values to the appropriate units.

First, we need to convert the given values into the appropriate units. - Temperature: Convert Celsius to Kelvin - \(T (K) = 25^{\circ} C + 273.15 = 298.15 K \) - Volume: Convert mL to L - \(V = 210 mL = 0.210 L \)
02

Rearrange the osmotic pressure formula to solve for n (moles of solute).

We can rearrange the osmotic pressure formula to solve for n by multiplying both sides by V and then dividing by R and T: \(n = 蟺 * V / (R * T)\)
03

Calculate the number of moles of lysozyme (n).

Now, we can plug in the values we've converted and calculated into the formula: \(n = (0.953 \thinspace torr) * (0.210 \thinspace L) / (0.0821 \thinspace L \thinspace atm \thinspace mol^{-1} \thinspace K^{-1}) * (298.15 \thinspace K) \) Since we need the pressure to be in atm for the ideal gas constant R, we must convert torr to atm: 1 atm = 760 torr 0.953 torr * (1 atm / 760 torr) = 0.001253 atm Now, we can plug in the pressure in atm: \(n = (0.001253 \thinspace atm) * (0.210 \thinspace L) / (0.0821 \thinspace L \thinspace atm \thinspace mol^{-1} \thinspace K^{-1}) * (298.15 \thinspace K) \) \(n = 3.212 \times 10^{-5} \thinspace mol \)
04

Calculate the molar mass of lysozyme.

We can now calculate the molar mass (M) of lysozyme by dividing the given mass (m) by the number of moles (n) we just calculated: \(M = \frac{m}{n}\) \(M = \frac{0.150 \thinspace g}{3.212 \times 10^{-5} \thinspace mol} = 4671.64 \thinspace g \thinspace mol^{-1}\)
05

Round the answer to an appropriate number of significant figures.

Since the given values were all in three significant figures, we should round our answer to three significant figures as well: \(M 鈮� 4670 \thinspace g \thinspace mol^{-1}\) The molar mass of lysozyme is approximately 4670 g/mol.

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Most popular questions from this chapter

Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) \(0.75 \mathrm{~L}\) of solution that is \(12.0 \% \mathrm{KBr}\) by mass (the density of the solution is \(1.10 \mathrm{~g} / \mathrm{mL}\) ), (d) a \(0.150 \mathrm{M}\) solution of \(\mathrm{KBr}\) that contains just enough KBr to precipitate \(16.0 \mathrm{~g}\) of \(\mathrm{AgBr}\) from a solution containing \(0.480 \mathrm{~mol}\) of \(\mathrm{AgNO}_{3}\).

By referring to Figure 13.15, determine whether the addition of \(40.0 \mathrm{~g}\) of each of the following ionic solids to \(100 \mathrm{~g}\) of water at \(40^{\circ} \mathrm{C}\) will lead to a saturated solution: (a) \(\mathrm{NaNO}_{3}\), (b) \(\mathrm{KCl}_{\text {, }}\) (c) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\), (d) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\)

During a person's typical breathing cycle, the \(\mathrm{CO}_{2}\) concentration in the expired air rises to a peak of \(4.6 \%\) by volume. (a) Calculate the partial pressure of the \(\mathrm{CO}_{2}\) in the expired air at its peak, assuming 1 atm pressure and a body temperature of \(37^{\circ} \mathrm{C}\). (b) What is the molarity of the \(\mathrm{CO}_{2}\) in the expired air at its peak, assuming a body temperature of \(37^{\circ} \mathrm{C}\) ?

At \(20^{\circ} \mathrm{C}\), the vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is 75 torr, and that of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is 22 torr. Assume that benzene and toluene form an ideal solution. (a) What is the composition in mole fraction of a solution that has a vapor pressure of 35 torr at \(20^{\circ} \mathrm{C}\) ? (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?

A car owner who knows no chemistry has to put antifreeze in his car's radiator. The instructions recommend a mixture of \(30 \%\) ethylene glycol and \(70 \%\) water. Thinking he will improve his protection he uses pure ethylene glycol, which is a liquid at room temperature. He is saddened to find that the solution does not provide as much protection as he hoped. The pure ethylene glycol freezes solid in his radiator on a very cold day, while his neighbor, who did use the \(30 / 70\) mixture, has no problem. Suggest an explanation.
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