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Chapter 11: Problem 67

What mass of glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)\), a nonelectrolyte, must be dissolved in \(200.0 \mathrm{~g}\) water to give a solution with a freezing point of \(-1.50^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
14.84 g of glycerin must be dissolved in 200.0 g of water to give a solution with a freezing point of -1.50 °C.

Step by step solution

01

Calculate the change in freezing point

First, we need to calculate the change in freezing point (ΔTf). The freezing point of pure water is 0 °C, and we are given that the freezing point of the solution is -1.50 °C. Therefore, the change in freezing point can be found by: \(ΔT_f = \) Initial freezing point - Final freezing point \(ΔT_f = 0 - (-1.50)\) \(ΔT_f = 1.50 \mathrm{~°C}\)
02

Calculate the molality

Next, we use the freezing point depression equation while taking i = 1 (as glycerin is a nonelectrolyte): \(ΔT_f = K_f * \textrm{molality}\) \(1.50 = 1.86 * \textrm{molality}\) Now, we can solve for molality: \(\textrm{molality} = \frac{1.50}{1.86}\) \(\textrm{molality} = 0.806 \mathrm{~mol/kg}\)
03

Use molality to find the moles of glycerin

Now that we have found the molality, we can use it to find the moles of glycerin. Molality is defined as the moles of solute (glycerin) per kg of solvent (water). We know that, there are 200.0 g of water, which can be converted to kg: \(200.0 \mathrm{~g} = 0.200 \mathrm{~kg}\) Now, we can calculate the moles of glycerin: \(0.806 \mathrm{~mol/kg} * 0.200 \mathrm{~kg} = 0.1612 \mathrm{~mol}\)
04

Calculate the mass of glycerin

Finally, we can find the mass of glycerin by multiplying the moles of glycerin with the molar mass of glycerin. The molar mass of glycerin (C3H8O3) can be found by adding the molar masses of its elements: 3C + 8H + 3O: \(3(12.01) + 8(1.008) + 3(16.00) = 92.09 \mathrm{~g/mol}\) Now, we multiply the moles of glycerin by the molar mass: \(0.1612 \mathrm{~mol} * 92.09 \mathrm{~g/mol} = 14.84 \mathrm{~g}\) So, 14.84 g of glycerin must be dissolved in 200.0 g of water to give a solution with a freezing point of -1.50 °C.

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Most popular questions from this chapter

How would you prepare \(1.0 \mathrm{~L}\) of an aqueous solution of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) having an osmotic pressure of \(15 \mathrm{~atm}\) at a temperature of \(22^{\circ} \mathrm{C} ?\) Sucrose is a nonelectrolyte.

What is ion pairing?

An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of \(31.57 \% \mathrm{C}\) and \(5.30 \% \mathrm{H}\). The molar mass is determined by measuring the freezing- point depression of an aqueous solution. A freezing point of \(-5.20^{\circ} \mathrm{C}\) is recorded for a solution made by dissolving \(10.56 \mathrm{~g}\) of the compound in \(25.0 \mathrm{~g}\) water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

From the following: pure water solution of \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(m=0.01)\) in water solution of \(\mathrm{NaCl}(m=0.01)\) in water solution of \(\mathrm{CaCl}_{2}(m=0.01)\) in water choose the one with the a. highest freezing point. d. lowest boiling point. b. lowest freezing point. e. highest osmotic pressure. c. highest boiling point.

When pure methanol is mixed with water, the resulting solution feels warm. Would you expect this solution to be ideal? Explain.
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