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Chapter 4: Problem 60

You make 1.000 L of an aqueous solution that contains 35.0 \(\mathrm{g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right) .\) (a) What is the molarity of sucrose in this solution? (b) How many liters of water would you have to add to this solution to reduce the molarity you calculated in part (a) by a factor of two?

Short Answer

Expert verified
(a) The molarity of sucrose in the initial solution is \(0.1022 M\). (b) To reduce the molarity by a factor of two, \(1.000 L\) of water should be added to the solution.

Step by step solution

01

Find the molar mass of sucrose

To find the molar mass of sucrose (\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)), we need to sum the molar masses of all the individual atoms in a molecule of sucrose. Sucrose molecule has: - 12 carbon atoms (C) with molar mass 12.01 g/mol - 22 hydrogen atoms (H) with molar mass 1.01 g/mol - 11 oxygen atoms (O) with molar mass 16.00 g/mol Molar mass of sucrose = (12 × 12.01) + (22 × 1.01) + (11 × 16.00) = 144.12 + 22.22 + 176 = 342.34 g/mol.
02

Calculate the moles of sucrose

Now, we have the mass of sucrose (35.0 g) and the molar mass of sucrose (342.34 g/mol). We can find the number of moles using the formula: Moles of sucrose = mass of sucrose / molar mass of sucrose Moles of sucrose = 35.0 g / 342.34 g/mol = 0.1022 mol
03

Calculate the initial molarity of the solution

Using the moles of sucrose and the volume of the solution, we can find the molarity with the following formula: Molarity = moles of solute / volume of solution in liters Molarity = 0.1022 mol / 1.000 L = 0.1022 M So, the molarity of sucrose in the initial solution is 0.1022 M.
04

Calculate the volume of water to add to halve the molarity

Now we want to halve the initial molarity, which means we need to find a new volume of solution that would make the molarity 0.0511 M (half of 0.1022 M). We can use the formula: Molarity_new × Volume_new = Molarity_initial × Volume_initial Volume_new = Molarity_initial × Volume_initial / Molarity_new Volume_new = (0.1022 M × 1.000 L) / 0.0511 M = 2.000 L So, we need to have 2.000 L of solution to halve the initial molarity. Since we started with 1.000 L of solution, we need to add an additional 1.000 L of water to the solution. Answer: (a) The molarity of sucrose in the initial solution is 0.1022 M. (b) To reduce the molarity by a factor of two, 1.000 L of water should be added to the solution.

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