/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 90 An aqueous solution of \(10.00 \... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 90

An aqueous solution of \(10.00 \mathrm{~g}\) of catalase, an enzyme found in the liver, has a volume of \(1.00 \mathrm{~L}\) at \(27^{\circ} \mathrm{C}\). The solution's osmotic pressure at \(27^{\circ} \mathrm{C}\) is found to be \(0.745\) torr. Calculate the molar mass of catalase.

Short Answer

Expert verified
The molar mass of catalase is approximately \(2.52 * 10^5 \mathrm{~g/mol}\).

Step by step solution

01

Convert temperature to Kelvin and pressure to atm

First, we need to convert the given temperature of 27掳C to Kelvin. We do this by adding 273.15. T = 27 + 273.15 = 300.15 K Next, we need to convert the osmotic pressure from torr to atm. We use the conversion factor: 1 atm = 760 torr 螤 = 0.745 torr * (1 atm / 760 torr) = 0.0009796 atm
02

Rearrange the osmotic pressure formula to find the number of moles (n)

Now we will rearrange the osmotic pressure formula to find the number of moles (n): n = (螤 * V) / (R * T) Where V is the volume (1.00 L), R is the ideal gas constant (0.0821 L atm/mol K), and T is the temperature (300.15 K).
03

Calculate the number of moles (n) of catalase

Now plug in the values for 螤, V, R, and T to calculate the number of moles: n = (0.0009796 atm * 1.00 L) / (0.0821 L atm/mol K * 300.15 K) n 鈮� 3.97 * 10^-5 mol
04

Calculate the molar mass (M) of catalase

Now that we have the number of moles, we can find the molar mass (M) of catalase using the formula: M = mass / n Where mass is the mass of catalase (10.00 g) and n is the number of moles (3.97 * 10^-5 mol). M = 10.00 g / (3.97 * 10^-5 mol) 鈮� 2.52 * 10^5 g/mol The molar mass of catalase is approximately 2.52 * 10^5 g/mol.

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Most popular questions from this chapter

a. Calculate the freezing-point depression and osmotic pressure at \(25^{\circ} \mathrm{C}\) of an aqueous solution containing \(1.0 \mathrm{~g} / \mathrm{L}\) of a protein (molar mass \(=9.0 \times 10^{4} \mathrm{~g} / \mathrm{mol}\) ) if the density of the solution is \(1.0 \mathrm{~g} / \mathrm{cm}^{3}\). b. Considering your answer to part a, which colligative property, freezing- point depression or osmotic pressure, would be better used to determine the molar masses of large molecules? Explain.

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