# specific gravity problem

Submitted by ian610 on Mon, 11/22/2010 - 11:57

could you please explain to me how this is solved..

how many grams of water must be added to 200 mL of NaOH solution in order to have a solution with a specific gravity of 1.157, 13.55%? (specific gravity of NaOH is 1.32, 28.83%)..

Not entirely sure. We need to take care with % value which are mass/volume of water, and SG values which are mass of the solution/mass of same volume of water.

If NaOH = 28.83 % it contains 28.83g/100g (or 100 mL) of water and has a combined mass of 128.83g. (Density of water = 1g/mL)

If SG = 1.32 then 100 mL of solution has a mass of 132g and 200 mL of solution has a mass of 264g.
This is made of water and NaOH
We know that 128.83g of solution contains 28.83g of NaOH, so by proportion
264g of solution contains 57.85g NaOH. This would contain 264-57.85 = 206.15g of water

So this is the mass of NaOH in 200 mL of solution and this will be added to water to make the diluted solution.

Diluted solution = 13.55% or 13.55g/100g of water
Mass of NaOH/mass of water = 13.55%
57.85g/mass of water = 13.55%
Mass of water = 57.85 x (100/13.55) = 426.94g of water. In my reckoning, this is the total volume or  mass of water you need
As there was already 206.15g of water, you would need to add 426.94-206.15 = 220.79g (of mL) of water

I do not know if this is right and I have may have confused myself (as well as you)

i think i am.. but according to the answer key.. the supposed answer is 285.48 grams.. i just need to understand how did they arrived at such answer and at the same time understand the concept of the problem.. really need your help with this one..