he says the answer is 490 ml but i'm getting 50 L...
"what volume of water must be added to 10.0 mL of 5.0 M HCL to make a solution of .10M HCL?
"what volume of water must be added to 10.0 mL of 5.0 M HCL to make a solution of .10M HCL?
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your friend is right, you would use the formula M1 (initial molarity) x V1 (initial volume)= M2(final volume) x V2(final volume) in this instance the final volume (V2) is your variable so your equation is 10.0mL(5.0M)=0.1M(x)when you isolate the variable you find that x=500ml, but it asks what volume of water must be added so you subtract your initial 10mL to get 490mL
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Alright so molarity1*volume1=molarity2*volume2
V1=.010L
M1=5.0M
V2=?
M2=.10M
5.0*.010=.10*V2
V2=5.0*.010/.10=.5L
I get .5L because I was taught to always convert to liters. But just by moving the decimal back over 3 places to the right you get you answer in milliliters. So thus with those numbers the answer I get is 500ml which is close. How your "guy" got 490ml is beyond me..I would like to know how. But I believe my answer is correct so we will see what others say.
V1=.010L
M1=5.0M
V2=?
M2=.10M
5.0*.010=.10*V2
V2=5.0*.010/.10=.5L
I get .5L because I was taught to always convert to liters. But just by moving the decimal back over 3 places to the right you get you answer in milliliters. So thus with those numbers the answer I get is 500ml which is close. How your "guy" got 490ml is beyond me..I would like to know how. But I believe my answer is correct so we will see what others say.