Louie1104
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What is the molarity of a solution if 325ml of the solution contains 46.8 grams of NaHCO3?

Sagot :

Eduard22sly

Answer:

1.714 M

Explanation:

We'll begin by calculating the number of mole in 46.8 g of NaHCO₃. This can be obtained as follow:

Mass of NaHCO₃ = 46.8 g

Molar mass of NaHCO₃ = 23 + 1 + 12 + (3×16)

= 23 + 1 + 12 + 48

= 84 g/mol

Mole of NaHCO₃ =?

Mole = mass / molar mass

Mole of NaHCO₃ = 46.8 / 84

Mole of NaHCO₃ = 0.557 mole

Next, we shall convert 325 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

325 mL = 325 mL × 1 L / 1000 mL

325 mL = 0.325 L

Thus, 325 mL is equivalent to 0.325 L.

Finally, we shall determine the molarity of the solution. This can be obtained as shown below:

Mole of NaHCO₃ = 0.557 mole

Volume = 0.325 L

Molarity =?

Molarity = mole / Volume

Molarity = 0.557 / 0.325

Molarity = 1.714 M

Therefore the molarity of the solution is 1.714 M

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