Molality

A-R-Q · **Joined:** Sat, 20 Nov 2010 21:17:02 UTC **Posts:** 432

What is the molality of a 10.5% by mass glucose (C6H12O6) sol'n? The density of the sol'n is 1.03g/mL

Molality (I call it "Mal") = mol solute/kg solvent
My biggest issue for this question is with getting rid of the 'mL'. I don't know of any other conversion factor to get rid of it. In fact, from the "by mass" part, I got:

10.5 g C6H12O6
---------------
100 g soln

So I I multiply this by:

1.03 g soln
--------------
1 mL soln

I get rid of the "g soln" term, but I don't know how to get rid of the "mL soln" term..

WAIT! I think I got it:
the mass of the H2O = (100 - 10.5g) C6H12O6
I assume that the mass of C6H2O6 = 10.5g

so then Mal = (10.5g)/(100-10.5g)^-3 <-- ^-3 because it has to be in kg...
Is this correct?

Shadow · **Posted:** Sat, 28 Jan 2012 18:31:48 UTC

A-R-Q wrote:

What is the molality of a 10.5% by mass glucose (C6H12O6) sol'n? The density of the sol'n is 1.03g/mL

Molality (I call it "Mal") = mol solute/kg solvent
My biggest issue for this question is with getting rid of the 'mL'. I don't know of any other conversion factor to get rid of it. In fact, from the "by mass" part, I got:

10.5 g C6H12O6
---------------
100 g soln

So I I multiply this by:

1.03 g soln
--------------
1 mL soln

I get rid of the "g soln" term, but I don't know how to get rid of the "mL soln" term..

You should know that there is $1 \text{mol}/6.02\times 10^{23}$ particles, which means if you know the density of a particle...

A-R-Q · **Joined:** Sat, 20 Nov 2010 21:17:02 UTC **Posts:** 432

Shadow wrote:

A-R-Q wrote:

What is the molality of a 10.5% by mass glucose (C6H12O6) sol'n? The density of the sol'n is 1.03g/mL

Molality (I call it "Mal") = mol solute/kg solvent
My biggest issue for this question is with getting rid of the 'mL'. I don't know of any other conversion factor to get rid of it. In fact, from the "by mass" part, I got:

10.5 g C6H12O6
---------------
100 g soln

So I I multiply this by:

1.03 g soln
--------------
1 mL soln

I get rid of the "g soln" term, but I don't know how to get rid of the "mL soln" term..

You should know that there is $1 \text{mol}/6.02\times 10^{23}$ particles, which means if you know the density of a particle...

Wait please look at the opening post again

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Molality

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