717goil g ileFalT FETwa a density of 124 gml at 20C A 45ml

717-go.il :g ::- -ile[FalT FE.Twa. a density of 1.24 g/ml at 20-C A 45ml sample of this sobution is mixed wi ihe volume of the final solution to 150 ml. Find the molarity of the Question 11 is worth 20 points The density of 11. A solution is prepared by dissolving 256 grams of manganese (II) dichromate in 475 mL. of water . the final solution is 1.56 g/ml. Calculate the concentration in the following units: molarity, molality. percent, mole fraction and mole percent.

Solution

10.

35% chloric acid means

In 100 mL of solution = 35 g of chloric acid

Or, in 1 mL of solution = (35/100) g of chloric acid

Or, in 45 mL of solution = 45 x (35/100) g of chloric acid

Or, in 45 mL of solution = 15.75 g of chloric acid

Molar mass of chloric acid = 84.5 g/mol

So, 84.5 g of chloric acid = 1 mol

Or, 1 g of chloric acid = (1/84.5) mol

Or, 15.75 g of chloric acid = (15.75/84.5) mol = 0.186 mol

Volume = 150 mL = 0.15 L

So, Molarity of chloric acid = 0.186 mol / 0.15 L = 1.24 M

11.

Manganese(II) Dichromate = MnCr2O7

Molar Mass of Manganese(II) Dichromate = 271 g/mol

So, 271 g of Manganese(II) Dichromate = 1 mol

Or, 1 g of Manganese(II) Dichromate = (1/271) mol

Or, 256 g of Manganese(II) Dichromate = (256/271) mol = 0.945 mol

Volume = 475 mL = 0.475 L

So, Molarity = 0.945 mol / 0.475 L = 1.99

Volume = 475 mL

Density = 1.56 g/mL

Mass = Volume x Density
         = 475 mL x 1.56 g/mL
         = 741 g

         = 0.741 Kg

Molality = 0.945 mol / 0.741 Kg = 1.28 m

Mass % = (256 g/ 741 g) x 100 = 0.345 %

Mole fraction = Moles of MnCr2O7 / Total moles
Moles of Manganese(II) Dichromate = 0.945 mol
Moles of water = 475 g / (18 g/mol) = 26.39 mol         (density of water = 1)

So, mole fraction = 0.945 mol / (0.945 mol + 26.39 mol)
                            = 0.945 mol / 27.335 mol
                            = 0.0346