How many grams of NaBr would be needed to prepare 750.00 mL of a 0.250 M NaBr solution?

Answer 1

Molarity, M = mol/L

M = 0.250 M

L = 0.75000 L (convert mL to L by dividing by 1000)

Rearrange to solve for mol:

mol = M * L = (0.250 M )(0.75000 L) = 0.1875 mol NaBr

Now we have the moles, we can convert to grams using the molar mass from the periodic table.

The molar mass of NaBr = (22.99 g/mol) + (79.90 g/mol) = 102.89 g/mol

(0.1875 mol) * (102.89 g/mol) = 19.3 g NaBr.

We record 3 sigfigs because we are limited by the 0.250 M which has 3 sigfigs.