Multiple choice

What amount of KOH (in g) is required for making 120 mL of a 0.10 M solution of potassium hydroxide in water?

  1. 56.1 g

  2. 0.012 g

  3. 0.67 g

  4. 1.34 g

  5. 2.68 g
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Correct answer. The amount of KOH required is (0.120 L) × (0.10 mol L−1) = 0.012 mol. The molar mass of KOH is 56.1 g. So, the weight of KOH required = 0.012 mol × 56.1 gmol−1 = 0.67 g.

Keep practicing — related questions