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What is the molality of a solution containing
10.0g Na2SO4 dissolved in 1000.0g of water?

1 Answer

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Answer:

Step-by-step explanation:

Use the definition of molality to guide you to find what is needed to solve for molality.

molality = mol of solute/kg of solvent

Step 1: Find molar mass of Na2SO4

Na: 2 x 23.00 g = 46.00 g

S: 1 x 32.06 g = 32.06 g

O: 4 x 16.00 g = 64.00 g

Molar mass of Na2SO4 = 142.06 g/mol

Step 2: Convert 10.0 g Na2SO4 into moles using molar mass of Na2SO4 from step 1.

10.0 g x (1 mol/142.06 g) = 0.070393 mol

Note that grams cancel and you are left with moles.

Step 3: Convert 1000.0 g of water into kilograms of water. Remember that 1000 g = 1 kg.

1000.0 g x (1 kg/1000g) = 1.0000 kg water

Note that grams cancel and you are left with kilograms.

Step 3: Use the definition of molality to find molality for this solution. The solute is Na2SO4. The solvent is water.

Definition of molality = mol of solute/kg of solvent

molality = 0.070383 mol/1.0000 kg

molality = 0.070393 m

There are 3 significant figures in 10.0 g and 5 significant figures in 1000.0 g. Use the least number of significant figures when multiplying or dividing. In this case 10.0 g has the least number of significant figures. The final answer with the correct number of significant figures is 0.0704 m.

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