What volume (in liters) of a 1.772 M BaCl2 solution is needed to obtain 123 g of BaCl2?

Volume required = 0.327 L

Explanation:

Given data:

Volume in L = ?

Molarity of solution = 1.772 M

Mass of BaCl₂ = 123 g

Solution:

First of all we will calculate the number of moles of BaCl₂,

Number of moles = mass/molar mass

Number of moles = 123 g/ 208.23 g/mol

Number of moles = 0.58 mol

Now, given problem will solve by using molarity formula.

Molarity = number of moles / volume in L

1.772 M = 0.58 mol / Volume in L

Volume in L = 0.58 mol  / 1.772 M

Volume in L = 0.327 L