How many grams of fosforic acid (H3PO4) are required to prepare 500 mL of a 0.2 M solution?

Mass = 9.8 g

Explanation:

Given dа ta:

Molarity of solution = 0.2 M

Volume of solution = 500 mL

Number of grams of phosphoric acid = ?

Solution:

First of all we will convert the volume milliliter to litter.

500 mL * 1 L/1000 mL

0.5 L

Molarity = moles of solute / volume in litter

0.2 M = number of moles / volume in litter

Number of moles = 0.2 mol/L * 0.5 L

Number of moles = 0.1 mol

Number of grams:

Number of moles = mass/molar mass

Mass = number of moles * molar mass

Mass = 0.1 mol * 98 g/mol

Mass = 9.8 g