Pure ethanol has a concentration of 95%. Ethanol is used to disenfect surfaces at a concentration of 75%. How much pure ethanol should be combined with water to make a 500mL solution of 70% ethanol?
I don't remember how to solve this.
Interesting you asked for 70 % ethanol and not 75%
v= volume needed of 95% then 500-w is the amount of water needed
.95 (v) = .70 (500) (since the AMOUNT of ethanol is the same before and after)
.95v = 350
Solve for v
v= 368.42
So you'll need 500 - 368.42 water = 131.58 ml
~jc
+129847 Pure ethanol couldn't have a concentration of "95%"....if it's pure, it's 100% ethanol....!!!!
And pure water has a concentration of 0% ethanol .......so.....let the amount of pure ethanol to be mixed = x
And....the amount of pure water = 500 - x .......and we have....
100x + 0 (500 - x) = 70 (500)
100x = 35000 divide both sides by 100
x = 350 ml of pure ethanol
And the amount of water = 500 - 350 = 150 ml
Interesting you asked for 70 % ethanol and not 75%
v= volume needed of 95% then 500-w is the amount of water needed
.95 (v) = .70 (500) (since the AMOUNT of ethanol is the same before and after)
.95v = 350
Solve for v
v= 368.42
So you'll need 500 - 368.42 water = 131.58 ml
~jc
