I. An experiment calls for 64.0 g of a 7% (wt.) aqueous solution of calcium nitr
ID: 1038789 • Letter: I
Question
I. An experiment calls for 64.0 g of a 7% (wt.) aqueous solution of calcium nitrate. How would you make up a liter of this solution? 2. A 3.6 m solution of calcium chloride, CaCh, is used in tractor tires to give them weight. The addition of CaClh also prevents water in the tires from freezing at temperatures above-20°C. What are the mole fractions of CaCh and water in such a solution? 3. A solution contains 8.89 x 103 mole fraction of I2 dissolved in 0.9911 mole fraction of CH2C2 (methylene chloride). What is the molality of 12 in this solution? 4. A 2.3m aqueous solution of citric acid, HCsH 0, is prepared, and has a density of 1.16 g/mL. What is this solution's molarity (M)?Explanation / Answer
1.
It means that,
Mass of calcium nitrate = 64 * 7 / 100 = 4.48 g.
Mass of water = 64 - 4.48 = 59.52 g.
SInce density of water = 1 g/mL
Volmue of water = 59.52 mL
Therefore,
59.52 mL of solution contains 4.48 g. of calcium ntrate
Then, 1000 mL of solution contains 1000 * 4.48 / 59.52 = 75.3 g. of calcium nitrate.
SO, take 75.3 g. of calcium nitrate and add minimum amount of water to dissolve all calcium nitrate. And then add the water till 1000 mL mark to get required solution.
2.
Moles of CaCl2 = 3.5 mol
Mass of water = 1000 g.
Moles of water = mass / molar mass = 1000 / 18.0 = 55.6 mol
Mole fraction of CaCl2 = 3.5 / (55.6 + 3.5) = 0.0592
Mole fraction of water = 55.6 / (55.6 + 3.5) = 0.941