To calculate molarity, divide the number of moles of solute by the volume of the solution in liters. The molarity of the glucose solution is 1.563 M. The hydronium ion concentration in a solution with 0.001 M hydroxide is 1 x 10^-11 M.
To calculate the molarity of a solution, we need to know the amount of solute and the volume of the solution. For Exercise 13.6.1 A, given 66.2 g of glucose (C6H12O6) dissolved to make 235 mL of solution, we follow these steps:
- Determine the molar mass of glucose, which is approximately 180.16 g/mol.
- Calculate the number of moles of glucose:
(66.2 g) / (180.16 g/mol) = 0.367 moles.
- Convert the volume from mL to L: 235 mL = 0.235 L.
- Divide the number of moles by the volume in liters to get the molarity: (0.367 moles) / (0.235 L) = 1.563 M.
The molarity of the glucose solution is 1.563 M.
For Exercise 17.3.2, 15.1.2, and those similar, we use the relationship between hydroniumion concentration and hydroxideion concentration given by the water ion product (Kw = 1 x 10-14 at 25 °C).
If the hydroxide ion concentration is 0.001 M, then:
[H3O+] = Kw / [OH-] = (1 x 10-14) / (0.001 M) = 1 x 10-11 M.
The hydronium ion concentration is 1 x 10-11 M in the solution.