2. Occasionally, solubility experiments are performed on a smaller scale, in ord

Question

2. Occasionally, solubility experiments are performed on a smaller scale, in order to use a minimal amount of chemicals. In these cases, calculated solubility limits can be "scaled up" to the proper 100 gram (or milliliter) amount of solvent by multiplying both the numerator and the denominator by the same value. Using this information, "scale up" a solubility limit of 4.25 g/5.0 mL. H:O to calculate the solubility of g KNO,/100 ml. H:C. 3. A student dissolves 1.50 grams of KNO, in 5.0 grams of water. Calculate the mass percent of the resulting solution. 4. A student dissolves 1.50 grams of KNO, in enough water to produce 6.2 ml. of solution. Calculate the molarity of the resulting solution.

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Answers

2. In 5 mL H2O 4.25 g KNO3 is soluble. Hence 100 ml H2O can solubilise (4.25*100)/5 g KNO3 = 85 g KNO3

3. 1.5 g KNO3 is present in 5 g H2O. Total mass = (1.5 + 5) g = 6.5 g. So, mass percent of KNO3 = (mass of KNO3/Total mass) * 100 = (1.5/6.5)*100 = 23.08

4. Molar mass of KNO3 = 101.1032 g/mol. Hence 1.5 g KNO3 means 1.5/101.1032 = 0.0148 mol. This is dissolved in 6.2 ml water. Now, Molarity = (mol content of solute/volume of solvent)*1000 = (0.0148*1000)/6.2 = 2.387. Therefore it's a 2.4 (M) solution.

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