Molality Question

[Ranelar]

Possibly dumb question, but I don't know the answer...

The problem is: What is the molality of 85% phosphoric acid, H3PO4, if the density of the solution is 1.70 g/ml.

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

I get that 1000g/kg * 1ml/1.7g = ml/kg
I get that 85/100 = 85% of the solution is phosphoric acid
I get that 1.70g/ml * 1 mol/98g = mol/ml

Put them together and you get mol/kg of 85% phosphoric acid: molality

What's the 100/15 for? 100g solution divided by 15g solvent? Why?

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[license2kill]

Where did you get this problem?

 

[Zerconia2921]

Possibly dumb question, but I don't know the answer...

The problem is: What is the molality of 85% phosphoric acid, H3PO4, if the density of the solution is 1.70 g/ml.

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

I get that 1000g/kg * 1ml/1.7g = ml/kg
I get that 85/100 = 85% of the solution is phosphoric acid
I get that 1.70g/ml * 1 mol/98g = mol/ml

Put them together and you get mol/kg of 85% phosphoric acid: molality

What's the 100/15 for? 100g solution divided by 15g solvent? Why?

Hey I dont know if i did this right but you get 5.6m and I got 5.1m i might be missing somthing but this is how i did it made sense to me.

Density = 1.70g/ml solution therefore 1.70g = .0017kg solution

.85g H3PO4 x 1mol/98gH3PO4 = 0.0086 moles

0.0086moles/0.0017kg = 5.1 moal

 

[Ranelar]

Hey I dont know if i did this right but you get 5.6m and I got 5.1m i might be missing somthing but this is how i did it made sense to me.

Density = 1.70g/ml solution therefore 1.70g = .0017kg solution

.85g H3PO4 x 1mol/98gH3PO4 = 0.0086 moles

0.0086moles/0.0017kg = 5.1 moal

yea, that's what I would intuitively do as well, but for some reason there's that 100/15 thing in the answer. I got this from DAT Achiever. It doesn't ask for a real answer btw, the answers are all in the form of unsolved equations like the one I posted.

 

[Ranelar]

I got (85/98)*10 as the answer. Where did you get this problem?

If I am wrong, then the answer is derived by dividing you answer by (15/100) whish is the same as mulitplying by the reciprocal.

I don't really get what you're saying here

 

[license2kill]

Hey I dont know if i did this right but you get 5.6m and I got 5.1m i might be missing somthing but this is how i did it made sense to me.

Density = 1.70g/ml solution therefore 1.70g = .0017kg solution

.85g H3PO4 x 1mol/98gH3PO4 = 0.0086 moles

0.0086moles/0.0017kg = 5.1 moal

How did you reason .0017kg solution from the density of the solution? Just curious...

 

[Zerconia2921]

How did you reason .0017kg solution from the density of the solution? Just curious...

Yea i used the density of the solution.
Let me re-think this its 2:44 am I got the time.

85 solute find moles of solute = 0.0086 moles

85% h3po4 rest 15% solvent
Density is 1.70g/ml solution

100g solution x1.70g solution = 170g = .170kg of solution and 85 percent of h3po4 gets dissociated then .170kg X.85 = .1445kg therefore .170kg - .1445kg = 0.0255 kg solvent

Molality = moles of solute / kg of solvent so 0.0086moles /0.0255kg solvent = 0.3 m

I dont know if this is right but i put in my 4 cents now.

 

[license2kill]

Its 4AM in Denver-I've got plenty of time.^^

This will be my last attempt. Here it goes, tell me if I am wrong, but I am feeling good about this one.
So the problem is asking for molality (mol of solute/kg of solvent).

I first use the density of the solution to multiply by 85/100 so that I get density of the acid only. (1.7*85)/100 with the same units of (g/ml).

Next, I take this density of solute only and divide by the molar mass of H3PO4 so that I get (1.7*85)/(100*98) in units of (mol solute/ml).--This makes up the numerator of the molality equation (mol of solute).

Now, we will work for the denominator of the molality equation, finding (kg of solvent. This can be done by following a similar pattern as the numerator of the molality equation. By multiplying the density of the solution by (15/100), we will actually come up with the density of the solvent only which comes out to be (1.7*15)/(100) in the same units of (g/ml). Now, since we want the denominator to be in kg, we take this new found density of solvent only and multiply by (1kg/1000g) yielding (1.7*15)/(100*1000) in units of (kg/ml). Now we have found the denominator of the molality equation.

If we bring these two components together, we can see that the ml units cancel out and get the answer that the book gave in units of (mole of solute/kg of solvent).

This reminds me of the good old days of gen chem, did you guys ever have CAPA homework for gen chem?