1. Five students determined the solubility of KNO,in water at 30°C g KNOs/ 100 g
ID: 715184 • Letter: 1
Question
1. Five students determined the solubility of KNO,in water at 30°C g KNOs/ 100 g HO 44.9 45.7 46.1 45.8 46.0 a. Determine the mean (average) of the data set. b. In preparation for a different solubility determination, a student measured the mass of an empty beaker (25.62 g) on a balance that could be read to t0.01 g. Then, KNO was added to the beaker and the mass recorded again as 35.18 g. Finally, water was added to the flask and the mass determined to be 50.03 g. The temperature of the mixture was increased until all of the solid had just dissolved. What was the solubility at this final temperature? 2. The solubility of O2 (g) in water is 4.43 mg O2/100 g H2O (20 "C and 1 atm). What is the molarity (M) of this saturated solution? 3, Can an aqueous solution be prepared that is 30% by mass KNO3 at 20°C? Explain. (Hint: Look up % by mass calculation if unsure and use the data from the literature values of the solubility of potassium nitrate- see pg 2)Explanation / Answer
1.(a) Given :- The data set of solubility of KNO3 in water at 30o C by 5 students
To find :- The mean (average) of the data set.
The formula to find mean = Sum of all the observations / Total no. of observations
Putting the given values from the table in the formula ,
Mean = (44.9 + 45.7 + 46.1 + 45.8 + 46.0) / 5
= 228.5 / 5
= 45.7
Hence , the mean (average) of the given set is 45.7.
1.(b) The mass of an empty beaker is given as 25.62 g.
After adding KNO3 , the mass becomes 35.18 g. So , the mass of KNO3 added is 35.18 - 25.62 g = 9.56 g.
After the water is added , the mass becomes 50.03 g. Hence , the mass of added water = 50.03 - 35.18 g = 14.85 g
To calculate solubility in g/100g H2O , the formula used is
Solubility = (Mass of solute dissolved * 100) / Mass of solvent
Putting the calculated values in the formula ,
Solubility = (9.56 * 100) / 14.85
= 64.3
Hence , the solubility at the final temperature is 64.3.
2. First of all , convert the solubility in standard unit which is g/100g . So , divide 4.43 mg/100g by 10 and hence , the solubility is 0,443 g/100g. Now , to find the molarity , we have to calculate the molar mass of the dissolved compound as the sum of mass of all atoms in the molecule. Atomic weights of corresponding elements are given in the periodic table of the chemical elements. In the example, it would be: Molar mass O2 = 2 * M(O) = 2 * 16 = 32 g/mole.
Divide the mass of the dissolved compound by its molar mass to calculate the number of moles. In our example, this would be: Number of moles (O2) = 0.443g/32g/mole = 0.0138 moles
3. Pg 2 not given . Incomplete information
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