Answer:
Mass of barium nitrate needed = 25.74 g
Explanation:
Given data:
Molarity of solution = 0.197 M
Volume of solution = 500 mL (500 mL× 1 L /1000 mL = 0.5 L)
Mass of barium nitrate needed = ?
Solution:
Molarity = number of moles of solute / volume in L
0.197 M = number of moles of solute / 0.5 L
Number of moles of solute = 0.197 mol/L ×0.5 L
Number of moles of solute = 0.0985 mol
Mass of barium nitrate:
Mass = number of moles × molar mass
Mass = 0.0985 mol × 261.32 g/mol
Mass = 25.74 g
