part 1 You have 2.21 L of water which contains 25 mg/L of PO43–. What is...

part 1 You have 2.21 L of water which contains 25 mg/L of PO43–. What is the total amount of phosphate in the sample? in g -----

part2 The label on a bottle of "organic" liquid fertilizer concentrate states that it contains 8.40 grams of phosphate per 100.0 mL and that 16.0 fluid ounces should be diluted with water to make 32.0 gallons of fertilizer to be applied to growing plants. What version of the dilution equation can be used to calculate the final phosphate concentration? Express your answer in terms of Vi, Ci, and Vf. Cf= Please Help

Expert Solution

Given

2.21 L of solution

25 mg/L of PO₄^3-

amount of phophate present in solution = Volume * concentration = 2.21 L * 25 mg/L = 55.25 mg = 0.05525 g

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part 2

Given

8.4 g of phophate per 100ml Ci = 8.4 g/100ml = 84 g/1000 ml = 84 g/L = Ci

volume of this solution = 16 ounces = 0.4732 L = 473.2 ml = Vi = 0.4732 L = Vi

so amount of phosphate in this 16 ounces of solution = (8.4 g/100ml)* 473.2 ml = 39.75 g

Vf = 32 gallons = 121.133 L

Vi*Ci = Vf * Cf

84 g/L * 0.4732 L = 121.133 L * Cf

Cf = 0.3281 g/L Answer

queen_honey_blossom answered 2 years ago

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