Question

calculate the molality of a solution containing 5.3 g of anhydrous Na2CO3 in 400 g of water?

Answer 1

Answer:

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 => 0.125 m.

Answer 2

The answer is 0.125 moles/kg.

Given:

5.3 g of anhydrous $Na_2CO_3$

400 g of water

To Find:

Molality of solution

Solution:

The formula of molality is

$m=\frac{moles}{mass of solvent}$

The molar mass of $Na_2CO_3$ is 106.

Mass of solvent = 0.4 kg

Hence

$m=\frac{5.3}{0.4*106} \\m=0.125$

The molality is 0.125 moles/kg.

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