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March 30, 2015

March 30, 2015

Posted by **Emma** on Wednesday, December 9, 2009 at 12:38am.

- Chemistry -
**Dr Russ**, Wednesday, December 9, 2009 at 10:04amThe question is written in a very odd way. There is no need add more NaCl to make the 0.45% drinkable as it is already drinkable.

However, in the spirit of the question, we need to add v ml of the 3.5% solution (=sea water!) to 1 liter and end up with 0.9%.NaCl. So the starting solution contains 4.5 g NaCl (I am assuming that the % is w/volume) we need to end up with

(1000+0.0045)+(vx0.035)g salt

in 1000+v ml

and these two divided x 100/1 = 0.9%

so

[(1000+0.0045)+(vx0.035)]x100/(1000+v)=0.9%

solve for v

I got 4.5 = 8.965V+0.009V^2

if we assume that v is small

I got v=8.965/4.5 = 1.99 ml

But check my maths.

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