Answer:
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
