Sodium monofluorophosphate (Na2PO3F) (144.0 g/mol) is the fluoride component in some modern toothpastes. What is the molality of a solution containing
a. 1.85 kg of H2O and 1.00 mol of sodium monoflourophosphate?
b. 125.0 g of H2O and 0.356 g of sodium monoflourophosphate?
c. 500.0 g of solution and 12.0 g of sodium monoflourophosphate?
- Roger the MoleLv 71 month agoFavorite Answer
(1.00 mol Na2PO3F) / (1.85 kg) = 0.54054 mol/kg = 0.541 m
(0.356 g Na2PO3F) / (144.0 g Na2PO3F/mol) / (0.1250 kg) = 0.019778 mol/kg =
(12.0 g Na2PO3F) / (144.0 g Na2PO3F/mol) / (0.5000 kg - 0.0120 kg) =
0.170765 mol/kg = 0.171 m
- 1 month ago
molality is (moles of solute/kg of solvent).
a) (1.00 mol Na2PO3F/1.85kg)= 0.541 molal
For b, convert g of H20 (the solvent) into kg. Then convert 0.356g of Na2PO3F into moles of Na2PO3F using the compound's molar mass (144.0 g/mol) and dimensional analysis. Plug in the numbers into the equation above for you answer.
For step c, repeat step b. Remember that solute + solvent is a solution. That affects what you do with the 500g.
Hope that helped, -Nick