GChem molality problem

[nartnad]

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(Achiever) What is the molality of 85% H3PO4 if the density of the solution is 1.7g/mL?

answer:

1000*(1/1.70)*(100/15)*(1.70/1)*(85/100)*(1/98)m

Any takers?

 

[nze82]

(Achiever) What is the molality of 85% H3PO4 if the density of the solution is 1.7g/mL?

answer:

1000*(1/1.70)*(100/15)*(1.70/1)*(85/100)*(1/98)m

Any takers?

>>An 85% solution of H3PO4 means 85g H3PO4 and 15g solvent for every 100g of the solution.

>>Molality = #moles/Kg solvent

a)Calculate the # of moles of H3PO4:

(1.7g solution/1ml solution)x(85g H3PO4/100g solution)x(1mole H3PO4/98g H3PO4)

b)Calculate kg of solvent:

(1.7g solution/1ml solution)x(15g solvent/100g solution)x(1Kg solvent/1000g)

c) Divide (a) by (b) and you get the answer.