Question

17, How many liters of saline solution (0.6%) can be made from 12 g of sodium chloride? 18. How many grams of boric acid crystals are needed to prepare 250 ml of 8% boric acid solution?

Answer 1

17. The weight of available sodium chloride is 12grams.

The weight of sodium chloride present in 1 litre of 0.6% saline solution is 6 grams.

The calculation is 0.6 % means 0.6 grams in 100ml (w/v%). So 1000 ml contains 6 grams.

As 6 grams are required for 1 litre of 0.6% saline solution, the amount of 0.6% saline solution that can be prepared for 12 grams is 2 litres.

18. The required concentration of boric acid solution is 8%

The required volume of solution is 250 ml.

8% solution means 8 grams of boric acid in 100 ml of solution (w/v%).

So 250 ml solution requires 20 grams of boric acid crystals.

The calculation is : weight of boric acid required for 250 ml is (8*250)/100 grams

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