how would you prepare 300.0 ml of a 0.45 M NaCl solution from solid NaCl and water
and from 1.0 M Nacl solution?

Question

anonymous · Answer

molecular mass of NaCl is 58.44 gm/mol . It means if you take 58.44 grams in 1 L of say water , it makes up to 1 M solution . So now answering your question,$$x= 0.45/1 * 300/1000 * 58.44$$ will fetch your desired concentration in 300 ml of water. so if you calculate it will be 7.8894 gm. So weigh 7.8894 gm (be careful about the digits that follow the decimal point, can be weighed only in analytical balance) and take in 300 ml water and dissolve it.

If you already have 1 M solution of NaCl, then use this formula

V1* M1= V2 * M2
lets say you have got 1 L of 1M NaCl solution with you
So V1 = x ml, M1 = 1 M
     V2= 300 ml, M2= .45 M
 $$V1= V2*M2/M1$$    
so the answer is ; 135 ml of 1 M NaCl , dilute to 300 ml with water. there you go....

All the best
Ghajesh