Topic: Molarity (Read 4714 times)

semistry · « **on:** September 25, 2005, 08:47:21 PM »

"How much water would be needed to dilute 2.00L of 6.00L M sulfuric acid until the concentration is 2.50 M?"

for this problem I got 0.883L. with this, i found the molar mass of H₂SO₄ to be 98.0734 g/mol... coverting to get moles/volume to get Molarity.

saibot · « **Reply #1 on:** September 25, 2005, 09:21:42 PM »

Here you can use the equation M1 x V1 = M2 X V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

rsixtyone · « **Reply #2 on:** September 25, 2005, 09:25:16 PM »

i attempted that and i got 0.833L

saibot · « **Reply #3 on:** September 25, 2005, 09:31:58 PM »

I'm getting 2.8 L.

M1 X V1 = M2 X V2
(6.00 M)(2.00L)=(2.50M)V2
(12.0)/(2.50)=V2
4.80=V2

The final volume of the 2.50 M solution is 4.80L. So the change in volume is:

Vf - Vi = deltaV
(4.80L) - (2.00L) = deltaV
2.80L = deltaV

So, you would have to add 2.80L to the original solution to make 4.80L of the final solution.

Borek · « **Reply #4 on:** September 26, 2005, 03:22:52 AM »

Quote from: saibot on September 25, 2005, 09:31:58 PM

So, you would have to add 2.80L to the original solution to make 4.80L of the final solution.

2.84L if you take density changes into account.

CASC rulez

Topic: Molarity (Read 4714 times)

semistry

Molarity

saibot

Re:Molarity

rsixtyone

Re:Molarity

saibot

Re:Molarity

Borek

Re:Molarity

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