How many mL would it take to prepare 500mL of a 0.10M Nitric Acid (HNO3) solution from a stock solution that is 69.0% pure and has a specific gravity of 1.42g/mL?
Stacey Warren - Expert brainly.com
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I don't even have the slighest clue where to start on this one.
ok i dont know how to go about specific gravity... but Im guessing you use that to calculate the molarity then you can simply do M1V1=M2V2 where you would plug the molarities and one volume to get the other.. hope that helps
I think you have to use the purity and specific gravity to find the molarity, i'll see what i can come up with
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great good luck! and tell me what you find plz :)
Well solution specific gravity is 1.42g/mL, 69.0% pure HNO3, so HNO3 specific gravity is .9798g/mL. Need molarity of HNO3 now
Oh wait.. you can use the Molecular weight of HNO3 H=1 , N=14, O=16 (16x3=48)
add them all up you get 63 g/mole and for molarity you need moles/litre so what you do here is .9798 g/ml / 63g/mole the g's will cancel out and you will get your molarity in moles/ml just divide that by 1000 to get it in litres and then you can solve the Q !
so .9798 g/mL divided by 63g/mole = 64.29 moles/mL = .06429 Molarity? (moles/L)
so then (500mL)/0.10M x (needed amt)/.06429M = 320.
ok.. when i divided .9798 by 63 i got 0.0156 moles/ml then divided that by a 1000 I got 1.56E-5 (0.0000156) then took that did M1V1=M2V2
so (0.10M)(500)/1.56E-5 =3.205E6 ml but thats a very big number lol
your answer however gives a reasonable amount... maybe I screwed up my calculations..0_o
no depends on how you divided the .9798, i got 64.29
i thought mine was incorrect, but your number is too big, mine is reasonable but i thought my math was wrong :)
plus i think it's 320, then you add water to get 500mL