Gchem Question Achiever #59
Started by anonymuz
85% H3PO4 means that 100 g of solution has 85 g H3PO4 and 15 g solvent:
85 g H3PO4 * 1 mol H3PO4 / 98 g H3PO4 = mol H3PO4
15 g solvent * 1 kg solvent / 1000 g solvent = kg solvent
Molality is the first quantity divided by the second quantity, which comes out to the answer given. Notice that the 1.7s and 100s cancel.
What is the molality of 85% phosphoric acid, H3PO4, if the density of the solution is 1.7 g/ml?
Answer: 1000*(1/1.7)*(100/15)*(1.7/1)*(85/100)*(1/98)
molality is moles solute / kg solvent
85% H3PO4 solution means 85% is H3PO4, and 15% is the solvent.
So if the density is 1.7g solution/1mL solution then 1.7g/mL * (85%) = grams of H3PO4/ mL solution or up above its written as (1.7/1)*(85/100). Then you want the moles so (1/98g/mol) = moles
Now you want the kg of solvent. Density is 1.7g solution / mL solution but 15% is the solvent. So 0.15*(1.7/1) = g of solvent/1mL. Since molality is for 1 kilogram you do (15/100)*(1.7g/1mL)*(1kg/1000g)=kg solvent
Anyway to combine it into 1 long thing like they have above
(1.7/1)*(85/100)*(1/98g/mol) / (0.15)*(1.7g/1mL)*(1kg/1000g)
if we take the inverse of the bottom part of the fraction ( the (0.15)*(1.7g/1mL)*(1kg/1000g) ) it will move to the top of the fraction.
So (1.7/1)*(85/100)*(1/98g/mol) * (100/15)*(1mL/1.7g)*(1000g/1kg)
This question is asked here a lot. There are some other ways to solve it too.
Answer: 1000*(1/1.7)*(100/15)*(1.7/1)*(85/100)*(1/98)
molality is moles solute / kg solvent
85% H3PO4 solution means 85% is H3PO4, and 15% is the solvent.
So if the density is 1.7g solution/1mL solution then 1.7g/mL * (85%) = grams of H3PO4/ mL solution or up above its written as (1.7/1)*(85/100). Then you want the moles so (1/98g/mol) = moles
Now you want the kg of solvent. Density is 1.7g solution / mL solution but 15% is the solvent. So 0.15*(1.7/1) = g of solvent/1mL. Since molality is for 1 kilogram you do (15/100)*(1.7g/1mL)*(1kg/1000g)=kg solvent
Anyway to combine it into 1 long thing like they have above
(1.7/1)*(85/100)*(1/98g/mol) / (0.15)*(1.7g/1mL)*(1kg/1000g)
if we take the inverse of the bottom part of the fraction ( the (0.15)*(1.7g/1mL)*(1kg/1000g) ) it will move to the top of the fraction.
So (1.7/1)*(85/100)*(1/98g/mol) * (100/15)*(1mL/1.7g)*(1000g/1kg)
This question is asked here a lot. There are some other ways to solve it too.
Also, if that's the answer they want they should specify that it is 85% w/w (85 g H3PO4/100 g solution) as opposed to 85% w/v (85 g H3PO4/100 mL solution). In the first case, you don't need the density and you get the answer they have. In the second case, you would need the density, but it would give you a different answer:
85% w/v means 85 g H3PO4 / 100 mL solution. 100 mL solution weighs 170 g according to the density given. That gives 170 g solution - 85 g H3PO4 = 85 g solvent.
If you keep going and calculate the molality, you'd get the same numerator (mol H3PO4), but a different denominator (kg solvent) so the answer would be different and not the one given.
85% w/v means 85 g H3PO4 / 100 mL solution. 100 mL solution weighs 170 g according to the density given. That gives 170 g solution - 85 g H3PO4 = 85 g solvent.
If you keep going and calculate the molality, you'd get the same numerator (mol H3PO4), but a different denominator (kg solvent) so the answer would be different and not the one given.
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