Question

Calculate the molality of a solution containing 5.3 g of anhydrous Na₂CO₃ in 400g of water​

Answer 1

One mole of anhydrous Na2CO3 weighs 106 grams . So number of moles of a solution containing 5.3 g of anhydrous Na2CO3 =>

5.3/106 => 0.05. Mass of the solvent => 400 g =>0.4 kg. Molality of the solution => Number of moles of solute/Mass of the solvent => 0.05/0.4 => 0.125m.

hey mate here you go!!!!;-)

Answer 2

Here is ur answer:

$m = \frac{ \frac{ \frac{5.3}{106} }{400} }{1000} = \frac{0.05}{0.4} = 0.125m$

Hope this will help you ❤