How many grams of NaHCO3 are needed to prepare 250 mL of 0.50 M
NaHCO3?

The weight of sodium bicarbonate required to prepare 0.5 M, 250 ml solution has been 10.50 grams.

Molarity can be given as moles of solute per liter of solution. The moles can be defined as the weight of the compound with respect to the molecular weight.

Molarity =

For the solution of 0.5 M Sodium bicarbonate, the weight can be calculated as;

Molecular weight of Sodium bicarbonate = 84.007 g/mol

weight = 10.50 grams.

The weight of sodium bicarbonate required to prepare 0.5 M, 250 ml solution has been 10.50 grams.

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10.5g

Explanation:

First, let us calculate the number of mole of NaHCO3 present in the solution. This is illustrated below:

Volume = 250mL = 250/1000 = 0.25L

Molarity = 0.5M

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 0.5 x 0.25

Mole = 0.125 mole

Now, we shall be converting 0.125 mole of NaHCO3 to grams to obtain the desired result. This can be achieved by doing the following:

Molar Mass of NaHCO3 = 23 + 1 + 12 +(16x3) = 23 + 1 +12 +48 = 84g/mol

Number of mole of NaHCO3 = 0.125 mole

Mass of NaHCO3 =?

Mass = number of mole x molar Mass

Mass of NaHCO3 = 0.125 x 84

Mass of NaHCO3 = 10.5g

Therefore, 10.5g of NaHCO3 is needed.

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