2. In the BACKGROUND information, it was stated that CaF2 has solubility, at roo

Question

2. In the BACKGROUND information, it was stated that CaF2 has solubility, at room temperature, of 0.00160 g per 100 g of water. How many moles of CaFe is this? Assuming that the density of a saturated solution of CaF2 is 1.00 g/mL, how many moles of CaF2 will dissolve in exactly 1.00 L of solution? Using the solubility curve for K2Cr207 to the right, what is the maximum amount of grams of K2Cr207 that would dissolve in 250.0 mL of water at 55 °C? 3. 80 70 60 g 50 2 40 30 20 10 0 10 20 30 40 50 60 70 80 90 100 Temperature (C)

Explanation / Answer

2. Molar mass of CaF2 is 78.07 grams/ mole

So, 0.00160 grams of CaF2 = 0.00160 grams / 78.07 grams/ mole = 2.05 * 10-5 mole

Density of water = 1 grams/ mL. Mass of 1 L (or 1000 mL) water = 1000 mL * 1 grsm/ mL = 1000 grams

100 grams of water dissolves 2.05 * 10-5 mole of CaF2

So, 1000 grams of water dissolves 2.05 * 10-5 mole * (1000 g/100 g) CaF2 = 2.05 * 10-4 mole CaF2

3.

Let us assume the density of water = 1 .0 gram/ mL

250 mL of water has the mass = 250 mL * 1.0 g/ mL = 250 grams

From the solubility curve, we find that at 55 0C temperature, 100 grams of water dissolves 34 grams of K2Cr2O7

So, at that temperature 250 mL ( = 250 grams) water dissolves 34 grams * (250 grams / 100 grams) K2Cr2O7 = 85 grams K2Cr2O7

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