Question

QUICK HELP PLEASE! Patrick did an experiment to study the solubility of two substances. He poured 100 mL of water at 20 °C into each of two beakers labeled A and B. He put 60 g of Substance A in the beaker labeled A and 60 g of Substance B in the beaker labeled B. The solution in both beakers was stirred for 1 minute. The amount of substance left undissolved in the beakers was weighed. The experiment was repeated for different temperatures of water and the observations were recorded as shown. Experimental Observations Substance Mass of Undissolved Substance at Different Temperatures (gram) 20 °C 40 °C 60 °C 80 °C A 5 10 14 18 B 60 60 60 60 Part 1: Which, if any, substance is soluble in water? Part 2: Explain how the data helped you determine solubility for both substances for temperatures 20 °C to 80 °C.

Answer 1

The correct answers are as follows:
1. SUBSTANCE A IS SOLUBLE IN WATER WHILE SUBSTANCE B IS INSOLUBLE IN WATER.
Looking at the information given in the question about the two substance, you will discover that 60 g of both substances were initially put in the beaker. While the mass of undissolved substance for substance A is variable, that of substance B remain constant at 60 g, this shows that the substance is not soluble in water.

2. Looking at the two substances given in the question, you will observe that for substance A, the degree of solubility in water decreases as the temperature of the water increases. For instance, at 20 degree Celsius, the mass of undissolved substance is 5 and as the temperature increases to 40, 60 and 80 degree Celsius, the mass of undissolved substances also increase to 10, 14 and 18 respectively. This shows that, the higher the temperature, the less of the substance will dissolve. For substance B, the mass of undissolved substance remain constant at all temperatures, which shows that the substance is insoluble in water. Thus, the data given helps to determine the solubility of the substance by giving the amount of substances that are insoluble at each temperature.