25.00 mL of a 0.2466 M iron(III) chloride stock solution contains: A salt stock

Question

25.00 mL of a 0.2466 M iron(III) chloride stock solution contains:

A salt stock solution ([NaCl] = 2.54 M) has 23.76 mL removed and placed in another container. How many grams of sodium chloride are in the solution removed?

25.00 mL of an iron chloride stock solution (made by dissolving 100.0 g of iron(III) chloride in water and diluting to a final volume of 2.500 L) contains:

A salt stock solution ([NaCl] = 2.54 M) has 23.76 mL removed. How many grams of sodium chloride are in the solution removed?        

Explanation / Answer

A salt stock solution ([NaCl] = 2.54 M) has 23.76 mL removed and placed in another container. How many grams of sodium chloride are in the solution removed?

Weight of sodium chloride are in the solution = 23.76 x 2.54 x 58.44 / 1000 = 3.5268 gram of NaCl

25.00 mL of an iron chloride stock solution (made by dissolving 100.0 g of iron(III) chloride in water and diluting to a final volume of 2.500 L) contains:

100.0 g of iron(III) chloride in 2.5 Liter = 100 / 162.2 x 2.5 = 0.2466 M

(Molar mass of iron(III) chloride: 162.2 g/mol)

Concentration of stock solution = 0.2466M

Third question repetation of first question

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