chemistry lab Fesslback Procedure: . The given barium chloride solution is made

Question

chemistry lab

Fesslback Procedure: . The given barium chloride solution is made up to 100ml in a standard flask - 20mL of solution is pipetted into a 250 ml. beaker . About 5mL. 2N HCl is added and diluted to 150mL with distilled water The solution is heated to boiling and a hot solution of 4N H:SO. (10 I5ml) is added drop . The solution containing the precipitate is heated in a water bath for 5 minutes . The precipitate is allowed to stand for an hour . The clear solution is decanted through an ashless filter paper (whatman N .The by drop with constant stir o. 40) e precipitate is washed with hot distilled water to free sulphate ions. The particles adhering to the s policeiman Finally the precipitate is washed once again. ides of the beaker and glass rod are removed by a .The dried filter paper is folded and placed in a crucible which has been previousl weighed The filter paper with the precipitate is first incinerated on a Bunsen burner by a low flame and then transferred to an electric burner . The crucible is transferred to desiccator and cooled When cold, the crucible is weighed. Heating, cooling and weighing are repeated till concordant values are obtained. Calculation: Mass of crucible + lid-a g Mass of crucible + lid+ Bariumsulphate bg Mass of Bariumsulphate (b-a) g. 233.36 of barium sulphate contain 137.36 g of barium. Mass of barium in (b-a) g of Bariumsulphate- Therefore, Mass of Barium in the whole of the given solution Result: Mass of Barium in the whole of the given solution Cite this Simulator

Explanation / Answer

233.36 g of barium sulfate contains 137.36 g barium

Hence, 1 g barium sulfate contains 137.36 / 233.36 g barium

(b-a) g barium sulfate contains 137.36 / 233.36 x (b-a) g barium =0.59 (b-a)  g barium

Since initial solution is 20 ml and total volume is 100 ml ,

then mass of barium in the whole solution = 0.59 (b-a) x 100 / 20 g barium = 2.95 (b-a) g barium

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