What volume of water (in mL) is required to prepare a 3.000 M solution by dissolving 43.83 g of of NaCl. Select one:
a. 250.0 mL
b. 2500.0 mL
c. 2.50 mL
d. none of these
e. 25.0 mL

Option A. 250 mL

Explanation:

From the question given above, the following data were obtained:

Molarity of NaCl = 3 M

Mass of NaCl = 43.83 g

Volume of water =..?

Next, we shall determine the number of mole in 43.83 g of NaCl. This can be obtained as follow:

Mass of NaCl = 43.83 g

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mole of NaCl =?

Mole = mass/molar mass

Mole of NaCl = 43.83/58.5

Mole of NaCl = 0.749 mole

Next, we shall determine the volume of water required to prepare the solution as follow:

Mole of NaCl = 0.749 mole

Molarity of NaCl = 3 M

Volume of water =..?

Molarity = mole /Volume

3 = 0.749 /volume

Cross multiply

3 × volume = 0.749

Divide both side by 3

Volume = 0.749/3

Volume = 0.25 L

Finally, we shall convert 0.25 L to millilitres (mL). This can be obtained as follow:

Recall:

1 L = 1000 mL

Therefore,

0.25 L = 0.25 L × 1000 mL / 1 L

0.25 L = 250 mL

Therefore, 0.25 L is equivalent to 250 mL.

Thus, the volume of water needed to prepare the solution is 250 mL

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