Question

Determine the number of grams of sodium carbonate needed to prepare 100.0 ml of a 2.5 m solution

Answer 1

26.5 g

Step-by-step explanation:

First we convert 100.0 mL to L:

• 100.0 mL / 1000 = 0.100 L

Now we calculate how many moles of sodium carbonate are needed, using the definition of molarity:

• Molarity = moles / liters

• moles = molarity * liters

• 2.5 M * 0.100 L = 0.25 mol

Finally we convert 0.25 moles of sodium carbonate into grams, using its molar mass:

• 0.25 mol * 106 g/mol = 26.5 g