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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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