Question

How much barium chloride would be needed to make 250 ml of a solution having the same concentration of chloride?

Answer 1

Two solution of same concentration would mean that the molarity of both the solutions is equal. Therefore, first we need to calculate the molarity of NaCl solution.

Mass of NaCl = 3.78 g

Molar Mass of NaCl = 58.5 g/mol

Volume of NaCl Solution = 100 mL = (100 / 1000) L

(Molarity)NaCl = Mass of NaCl in grams / (Molar Mass of NaClxVolume of solution in L)

(Molarity)NaCl = (3.78 x 1000) / (58.5 x 100)

(Molarity)NaCl = 0.65 mol/L

Now, (Molarity)NaCl = (Molarity)BaCl2

Mass of BaCl2 = m g

Molar Mass of BaCl2 = 208.33 g/mol

Volume of BaCl2 solution = 250 mL = (250 / 1000) L

(Molarity)BaCl2 = Mass of BaCl2 in grams/(Molar Mass of BaCl2xVolume of solution in L)

0.65 = (m x 1000) / ( 208.33 x 250) g

m = (0.65 x 208.33) / 4 g

m= 33.8 g

Hence, 33.8 g of BaCl2 is required to make 250 ml of solution having same concentration of chloride as one containing 3.78g NaCl per 100ml.