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Ugo [173]
2 years ago
8

How many grams of nacl are required to make 250.0 ml of a 3.000 m solution?

Chemistry
1 answer:
ohaa [14]2 years ago
8 0
Molarity (M) = moles of solute (mol) / Volume of the solution (L)

Molarity of the solution = 3.000 M
Volume of the solution = 250.0 mL = 0.25 L
moles in 250.0 mL = molarity x volume of the solution
                              = 3.000 M x 0.25 L
                              = 0.75 mol

Hence, 0.75 mol of NaCl is needed to prepare 250.0 mL of 3.000 M NaCl solution.

Moles (mol) = mass (g) / molar mass (g/mol)

Moles of NaCl in 250.0 mL = 0.75 mol
Molar mass of NaCl           = <span>58.44 g/mol
Mass of NaCl in 250.0 mL = Moles x Molar mass
                                          = 0.75 mol x </span><span>58.44 g/mol
                                          = 43.83 g

Hence, 43.83 g of NaCl is needed to prepare 250.0 mL of 3.000 M solution.</span>
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Answer:

17.55 g of NaCl

Explanation:

The following data were obtained from the question:

Molarity = 3 M

Volume = 100.0 mL

Mass of NaCl =..?

Next, we shall convert 100.0 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

100 mL = 100/1000

100 mL = 0.1 L

Therefore, 100 mL is equivalent to 0.1 L.

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

Molarity = 3 M

Volume = 0.1 L

Mole of NaCl =?

Molarity = mole /Volume

3 = mole of NaCl /0.1

Cross multiply

Mole of NaCl = 3 × 0.1

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Mole of NaCl = 0.3 mole

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Cross multiply

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Therefore, 17.55 g of NaCl is needed to prepare the solution.

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