# Answer to Question #52905 in Inorganic Chemistry for suad

Question #52905

what is the molality of a 0.142M Na2PO4(aq) solution that has a density of 1.015 g/mL?

Expert's answer

Molality = Number of moles of solute / Mass of solvent in kgs

1,015 g/ml = 1,015 kg/L

0.142M means that in 1L of solution there are 0,142 moles of Na2PO4

So Molality = 0,142moles / 1.015kg = 0.1399 moles/kg

Answer: 0.1399 m

1,015 g/ml = 1,015 kg/L

0.142M means that in 1L of solution there are 0,142 moles of Na2PO4

So Molality = 0,142moles / 1.015kg = 0.1399 moles/kg

Answer: 0.1399 m

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