1a. A salt stock solution ([NaCl] = 2.54 M) has 23.76 mL removed. How many grams

Question

1a. A salt stock solution ([NaCl] = 2.54 M) has 23.76 mL removed. How many grams of sodium chloride are in the solution removed?

a. 3.53 g NaCl

b. 0.0604 g NaCl

1b. 25.00 mL of a 0.2466 M iron(III) chloride stock solution contains:

a. 6.165 * 10-3 g iron (III) chloride

b. 25.00 g iron (III) chloride

c. 1.000 g iron (III) chloride

1c. Calculate the concentration (in molarity) of iron cations in 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. This makes an iron(III) chloride stock solution whose Fe3+ concentration is:

a. 0.2466 M

b. 1.200 M

c. 40.00 M

1d. Using a 2.54 M NaCl stock solution, describe the process of creating a diluted solution of sodium chloride with [NaCl]= 2.75 M and final volume 2.500 L

a. Take 2.706 liters of stock and dilute to 2.500 L

b. Take 370.3 mL of stock and dilute to 2.500 L.

c. That solution can't be made as described because it's more concentrated than the stock solution you're trying to make it from.

Explanation / Answer

1a)

Molarity = moles / volume

2.54 = moles / 0.2376

moles = 0.0604

moles = mass / molar mass

0.0604 = mass / 58.5

mass of NaCl removed = 3.53 g NaCl

1b).

Molarity = moles / volume

0.2466 = moles / 0025

moles = 6.165 x 10^-3

mass of iron (III) chloride = 1.000 g

1c).

moles of iron(III) chloride = 100 / 162.2 = 0.6165

Molarity = 0.6165 / 2.5 = 0.2466

concentration of Fe+3 = 0.2466 M

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