You need to make an aqueous solution of 0.197 M barium nitrate for an experiment in lab, using a 500 mL volumetric flask. How many grams of solid barium nitrate should you add?

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

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