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How many grams of water would you add to $\pu{1.00 kg}$ of $\pu{1.44m}$ $\ce{CH3OH (aq)}$ to reduce the molality to $\pu{1.00 m}$ $\ce{CH3OH}$?

I have tried $M_1V_1=M_2V_2$. I have tried converting it to kg of $\ce{CH3OH}$ and subtracting it from the $\pu{1.00kg}$ to get kg of $\ce{H2O}$. All of which have worked to no avail.

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  • $\begingroup$ Do they say the density of $CH_3OH$? $\endgroup$ – Simon-Nail-It Nov 25 '14 at 2:25
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    $\begingroup$ I am a little confused with your units here. What do you mean with 1.44m? Is it an solution ethanol in water with a concentration of $1.44~\mathrm{\frac{mol}{L}}$? or is this a given molality? Or does it refer to a pure solution of ethanol and you have $1.44~\mathrm{mol}$ of it? Getting the units right would be the first step of finding the solution to your problem. $\endgroup$ – Martin - マーチン Nov 25 '14 at 2:33
  • $\begingroup$ As far as I am aware it is molality. So moles of solute per kg of solvent. There is also no given density."How many grams of water would you add to 1.00 kg of 1.44m CH3OH(aq) to reduce the molality to 1.00 m CH3OH?" Is the whole question that I am given. $\endgroup$ – JuliusDariusBelosarius Nov 25 '14 at 3:12
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If the molality is $1.44 \ce{m}$, it means that, there is 1.44 moles of methanol in $1 \ce{kg}$ of water.

To prepare this solution, we add the mass of $1.44\times 32= 46.08\ce{g}$ of methanol to $1000\ce{g}$ of water.

The mass of the final solution is $1000+46.08= 1046.08\ce{g} $.

Here, you have 1 kg of this solution (water + methanol). So, it contains $\frac {46.08 \times 1000}{1046.08}=44.05\ce{g}$ of methanol.

The mass of water in this one kg-solution is $1000-44.05=955.95\ce{g}$

The objective of the problem is to prepare a solution of molality $1 \ce{m}$,

i.e, it is prepared by adding $32\ce{g}$ of methaonl to $1000\ce{g}$ water.

It can also be prepared by adding $44.05\ce{g}$ of methanol to $\frac {44.05 \times 1000}{32}=1376.56 \ce{g}$ of water.

Our solution already contains $955.95\ce{g}$ of water.

So, we need to add only $1376.56-955.95=420.61\ce{g}$ of water.

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Firstly, 1.44 m of CH3OH yields a solution of 1.046 L, (where 1.44 m means (1.44*32=46.08) grams of CH3OH in 1 Kg of H2o, so solution weight will be 1.046 Kg) but we have a 1 Kg solution, so we say:


1.046 Kg solution contains 46.08 g of CH3OH


1.000 Kg solution contains x g of CH3OH


x=CH3OH grams in your solution = 46.08×1000/1046.08 = 44.05 g


and water weight in your solution is 1000−44.05=955.95 g


now, to prepare the new solution (1 m) we say:


1 kg of water contains 32 g of CH3OH


y Kg of water contains 44.05 g of CH3OH


y= weight of water required for new solution =44.05×1000/32=1376.56g


then 1376.56 - 955.95 = 420.6 g is water that should be added to the 1.44 m solution to have the 1 m solution .


Note: 1 Kg almost equals 1 L (for water)

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