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.

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## Molarity relates the moles

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