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01-04-2020
Chemistry

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6) What is the molality of 0.4 moles of KCl in 1500 g of solvent?

Respuesta :

09-04-2020

Answer:

Molality = 0.267 mol/Kg

Explanation:

Given data:

Moles of KCl = 0.4 mol

Mass of solvent = 1500 g

Molality of solution = ?

Solution:

Formula:

Molality = moles of solute / kg of solvent

Now we will convert the g into kg.

1500 g× 1kg/ 1000 g

1.5 kg

Now we will put the values in formula.

Molality = 0.4 mol/ 1.5 kg

Molality = 0.267 mol/Kg

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