Question

Describe how you would prepare 0.05mol/dm³ of sodium carbonate in a total of 250ml

Answer 1

Answer:

1.325 g of sodium carbonate will dissolve in 250 mL in order to prepare 0.05 M solution.

Explanation:

Given data:

Molarity of solution = 0.05 mol/L (dm³ = L)

Volume of solution = 250 mL  

Mass of sodium carbonate needed = ?

Solution:

First of all we will convert the volume in mL to L.

(250/1000 = 0.25 L)

Molarity = number of moles / volume in litter

0.05 mol/ L = n / 0.25 L

n = 0.05 × 0.25

n = 0.0125 mole

Mass of sodium carbonate = number of moles × molar mass

Mass of sodium carbonate = 0.0125 mol × 106 g/mol

Mass of sodium carbonate = 1.325 g

1.325 g of sodium carbonate will dissolve in 250 mL in order to prepare 0.05 M solution.