3.40 L of a 0.130 M CaCl2 solution

Question

Calculate the grams of solute needed to prepare

Answer 1

As in your previous question, we have 0.130 M CaCl₂ which mean 0.130 moles CaCl₂ per liter of solution.

Since we want to make 3.40 L of solution, we calculate total moles of CaCl₂ needed:

*** 3.40 L x 0.130 mo/L = 0.442 moles CaCl₂ needed.

Next, we convert this to grams using the molar mass of CaCl₂:

*** 0.442 moles CaCl₂ x 110.98 g/mol = 49.1 g CaCl₂ needed (to 3 significant figures)