What is the new concentration if you mix 8 g of solute in 1050 ml of solution and then add 400 ml of water?

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gsenviro | College Teacher | (Level 1) Educator Emeritus

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Let us assume that the solute is soluble in water and that the solution did not contain any solute in the beginning. In the first part, we added 8 g of solute in 1050 ml of solution. The concentration of the solution at this time can be calculated as the ratio of mass of solute and volume of solution. That is,

Concentration of solution = mass of solute/volume of solution = 8 gm/ 1050 ml = 0.00762 g/ml or, 7.62 gm/lt or 7.62 mg/ml.

When we add another 400 ml of water, the solution gets diluted. The solution now has a volume of 1450 ml (= 1050 ml + 400 ml) and the new concentration is now different, as compared to original solution and can be calculated as:

New concentration of solution = 8 gm/ (1050 + 400) ml = 0.00552 gm/lt or, 5.52 gm/lt or 5.52 mg/ml.

Hope this helps.