Question

Prepare a 1.00 L of a 0.25 M solution of sodium carbonate. Determine the amount of sodium carbonate needed in correct significant figures.

Answer 1

Answer:

Mass = 26.50 g

Explanation:

Given data:

Mass of sodium carbonate = ?

Molarity of solution = 0.25 M

Volume of solution = 1.00 L

Solution:

Molarity = number of moles / volume in litter

0.25 M = number of moles / 1.00 L

Number of moles = 0.25 M × 1.00 L

Number of moles = 0.25 mol

Mass:

Mass = number of moles × molar mass

Mass = 0.25 mol × 106 g/mol

Mass = 26.50 g