destroyer g-chem #136

[navneetdh]

I have the 2008 version so i am not sure if the question numbers will be same.

The question asks

at 0 degrees celsius a gas is dissolved in H2o at a mole fraction of 2x10^-18. What is the molality of the solution given the molecular wt of the gas is 32g/mol

a. 5x10^-4
b. 1x10^-14
c.1x10^-16
d.1x10^-12
e. 4x10^-7

I see the solution in the back but i have no clue why they assumed 1 mole of water????

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[UCB05]

I have the 2008 version so i am not sure if the question numbers will be same.

The question asks

at 0 degrees celsius a gas is dissolved in H2o at a mole fraction of 2x10^-18. What is the molality of the solution given the molecular wt of the gas is 32g/mol

a. 5x10^-4
b. 1x10^-14
c.1x10^-16
d.1x10^-12
e. 4x10^-7

I see the solution in the back but i have no clue why they assumed 1 mole of water????

it's an easy number to use to derive moles of solute and mass of solvent based on the given info. Since it's a molal calculation, it doesnt matter what number you choose to use because the mol/kg ratio calculation will remain the same. You could assume 1, 2, 10mols H20, or 1, 2, 10mols solute to get the same answer, but 1 mol of water makes things easy to calculate.

 

[nze82]

I have the 2008 version so i am not sure if the question numbers will be same.

The question asks

at 0 degrees celsius a gas is dissolved in H2o at a mole fraction of 2x10^-18. What is the molality of the solution given the molecular wt of the gas is 32g/mol

a. 5x10^-4
b. 1x10^-14
c.1x10^-16
d.1x10^-12
e. 4x10^-7

I see the solution in the back but i have no clue why they assumed 1 mole of water????

You're given the mole fraction of the gas, so let's start with that:

X = Mole gas / Mole total = 2E-18

Mole gas / (Mole gas) + (Moles solvent-Water) = 2E-18

Now, we need to choose a # for "Moles solvent." In this case they chose 1. Why?
First>>It makes solving for "Moles gas" that much easier using the above formula.

Second>>We're looking for molality of the gas (m = Mole gas/Kg Solvent). Now, choosing 1 mole of H2O (solvent) is ideal, because you should know 1 mole of water weighs 18g (18E-3 Kg), so you can easily use this number and solve for molality of your gas.

 

[navneetdh]

Thanks guys, that helped a lot.