could you please help me with this question:
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= 1.32, 28.83%).
the answer is 285.48 g. I just want to know the solution to this problem. thank you.
Molarity relates the moles per liter which means 1 liter so here we can find the molarity of the two solutions while we work on with stats given in the problem & 1 liter of each.
SG is 1.32g/ml so , as we have 1 liter solution conver it to ml so , 1000 ml as it is 1.32 g in 1 ml if we have 1000 ml we have
1320 grams of solution
28.83% of 1320 grams of solution = 380.556 grams of NaOH
convert mass to mole using the molar mass
380.556 grams/40.00 = 9.514 moles of NaOH
1000 ml =1 L
Molrity = 9.514 / 1 = 9.514 M NaOH solution .
similarly , we have 1157 grams of solution
13.55% of 1157 grams of solution = 156.774 grams of NaOH
using molar mass:
156.774 grams/ 40.00 g/mol = 3.919 moles of NaOH
so 3.919 Molar NaOH
now to your question, using the dilution formula:
M1V1 = M2V2
9.514 x 200ml = 3.919 x V2
V2 = 485.53 ml
the original solution must be diluted up to that volume in order to become 1.157, 13.55%,
Since you started with 200 ml you must have added 285.53 ml to it ....pay attention to correct number of sig fig .
Hope this helps