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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: semistry on September 25, 2005, 08:47:21 PM

Title: Molarity
Post by: 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.

Title: Re:Molarity
Post by: saibot 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.

Title: Re:Molarity
Post by: rsixtyone on September 25, 2005, 09:25:16 PM

i attempted that and i got 0.833L

Title: Re:Molarity
Post by: saibot 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.

Title: Re:Molarity
Post by: Borek 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 :)