The problem is: Describe how 1.0 L of 0.50 M NaClO could be created from a commercial bleach solution, which is 5.25% NaClO by mass and has a density of 0.953 g/mL? I get .000186 bleach when I do this, but this is obviously wrong... how do you do this problem?
I tried doing 1 L * (.5 mol/L) * (74.44 g/mol)(.0525) (this is molar mass) * (.953 mL/g) * (1 L/1000 mL) ... is this correct? The answer should be .74 bleach and .26 water but I didn't get that. How do I do it? Thanks.
You do it correctly
To make a 1.0L 0.5 mol sol'n of NaClO, you will need 37.22g of NaClO.
The Stock sol'n is 5.25% by mass, and density is 0.953g/ml,
so 5.25/100 = X/0.953 where X = grams of NaClO in a litre of stock solution.
thus ======> X = 50.0325g
Since we need 37.22g of NaClO in our desired solution, to determine the volume of stock solution, ===>37.22/X
Therefore 0.74L of stock sol'n will be required to make a 1.0L 0.5mol sol'n.
Montrealais wrote:
To make a 1.0L 0.5 mol sol'n of NaClO, you will need 37.22g of NaClO.
The Stock sol'n is 5.25% by mass, and density is 0.953g/ml,
so 5.25/100 = X/0.953 where X = grams of NaClO in a litre of stock solution.
thus ======> X = 50.0325g
Since we need 37.22g of NaClO in our desired solution, to determine the volume of stock solution, ===>37.22/X
Therefore 0.74L of stock sol'n will be required to make a 1.0L 0.5mol sol'n.
please posit your proof in english
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