an aqueous solution of copper sulphate was left out in the hot sun. d heat caused some of the d water to evaporate fr d solution. d original 150g solution had a concentration of 10% by mass. if d remaining solution has a concentration of 15% by mass, wat volume volume of water evaporated?
1) 10 ml 2) 15 ml
3) 50 ml 4) 100 ml
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Initially x gm of copper sulphate and 150 grams of solution are there.
x / 150 = 10 %
x = 150 * 10 /100 = 15 gm
Hence, original amount of water in the solution = 150 - 15 = 135 gms
After some water say y gms evaporated. So only (135 - y) gms os remaining. But copper sulphate remains at 15 gms in the solution.
15 / (15 + 135 - y) = 15% = 15 /100
150 - y = 100 => y = 50 gms
hence 50 ml of water has evaporated, as the density of water is 1 gm/ml.
x / 150 = 10 %
x = 150 * 10 /100 = 15 gm
Hence, original amount of water in the solution = 150 - 15 = 135 gms
After some water say y gms evaporated. So only (135 - y) gms os remaining. But copper sulphate remains at 15 gms in the solution.
15 / (15 + 135 - y) = 15% = 15 /100
150 - y = 100 => y = 50 gms
hence 50 ml of water has evaporated, as the density of water is 1 gm/ml.
kvnmurty:
thanks and u r welcom
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