Question

# Excess (NH4)2SO4 was added to a 70.0 mL solution containing BaCl2 (MW = 208.23 g/mol). The...

Excess (NH4)2SO4 was added to a 70.0 mL solution containing BaCl2 (MW = 208.23 g/mol). The resulting BaSO4 (MW = 233.43 g/mol) precipitate had a mass of 0.2790 g after it was filtered and dried. What is the molarity of BaCl2 in the solution?

Balanced chemical reaction is as follow

(NH4)2SO4 +   BaCl2  BaSO4 + (NH4)2Cl2

1 mole of BaCl2 produce 1 mole of   BaSO4

No. of mole = mass in gm/molar mass

No. of mole of BaSO4 = mass in gm of BaSO4/molar mass of BaSO4

= 0.2790/233.43 = 0.0011952 mole of BaSO4 priduced

1 mole of BaCl2 produce 1 mole of   BaSO4 then to produce 0.0011952 mole of BaSO4 0.0011952 mole of  BaCl2 required

thus, 70 ml solution contain 0.0011952 mole of BaCl2

70 ml = 0.070 L

molarity = no. of mole of solute/volume of solution in liter

0.0011952/0.070 = 0.0170 M

molarity of BaCl2 in the solution = 0.0170 M

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