4. For 6.7 g of sodium acetate and 6.7 g of magnesium acetate, calculate a) the

Question

4. For 6.7 g of sodium acetate and 6.7 g of magnesium acetate, calculate a) the moles of each salt and b) the moles of acetate ion in the sodium acetate and in the magnesium acetate.

5. If 6.7 g of sodium acetate or 6.7 g of magnesium acetate is dissolved in a total volume of 0.200 L of water calculate

a) the molarity of sodium acetate

b) the molarity of magnesium acetate

c) the molarity of acetate ion in each solution 6. How many grams of EDTA tetrahydrate are needed to prepare 0.100 L of a 0.020 M solution?

6. How many grams of EDTA tetrahydrate are needed to prepare 0.100 L of a 0.020 M solution?

Explanation / Answer

Sodium acetate = CH3COONa and Magnesium acetae = (CH3COO)2Mg

(4) (a) Number of moles of sodium acetate = 6.7/82 = 0.082 mol

Number of moles of magnesium acetate = 6.7/142.4 = 0.047mol

(b) Moles of acetate ion in sodium acetate = 0.082

Because each sodium acetate has one mole of acetate ions

moles of acetate ion in magnesium acetate = 2*0.047 = 0.094 mol

because each magnesium acetate has two moles of actate ions.

(5.) (a) Molarity = number of moles / volume of solution

molarity of sodium acetate = 0.082/0.200 = 0.41M

(b) Molarity of magnesium acetate = 0.047/0.200 = 0.235 M

(c) Molarity of acetate in sodium acetate = 0.41M

and molarity of acetate in magnesium acetate = 2*0.235 = 0.47 M

(6) Molarity = Number of moles / volume of solution

0.020 = n / 0.100

n = 0.002 mol

But number of moles = mass / molar mass

0.002 = mass / 452

mass = 0.904 g. of EDTA tetrahydrate are needed.

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