Chemistry help!!!?

Question

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? 

Answer 1

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a.

(1.00 mol Na2PO3F) / (1.85 kg) = 0.54054 mol/kg = 0.541 m

b.

(0.356 g Na2PO3F) / (144.0 g Na2PO3F/mol) / (0.1250 kg) = 0.019778 mol/kg =

0.0198 m

c.

(12.0 g Na2PO3F) / (144.0 g Na2PO3F/mol) / (0.5000 kg - 0.0120 kg) =

0.170765 mol/kg = 0.171 m

Answer 2

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