Tuesday

January 27, 2015

January 27, 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.

**Answer this Question**

**Related Questions**

Math - How would you go about making 250ml of a 20X stock solution of NaCl to ...

Math(Please help) Thank you - How would you go about making 250ml of a 20X stock...

chemistry - The solubility of NaCl in water is 35.7 g NaCl/100 g H2O. Suppose ...

chemistry - The solubility of NaCl in water is 35.7g NaCl/100g H2O. Suppose that...

chemistry - The solubility of NaCl in water is 35.7g NaCl/100g H2O. Suppose that...

bio - You have a sample of NaCl that contains 100 mg/ml NaCl. You are trying to ...

chemistry - In a laboratory experiment, a 18.0 ml sample of NaCl solution is ...

science - In the lab you need to make 1.00 L of a 0.300 M NaCl solution. You had...

science - answer to water conductance lab on conductivity of solutionof 500ppm ...

chemistry - Making the simplistic assumption that the dissolved NaCl(s) does not...