Asked by julie

What mass of solute is present in each aqueous solution?
40 mL of 6.0 mol/L, H2SO4, solution

Soo I did:
C=n/V

40 mL= 0.04 L

C= 6.0 mol/L/0.04 L
= 150 mol/L

MH2SO4= 2.02 g/mol + 32.07 g/mol + 64 g/mol= 98.09 g/mol

mass H2SO4= 150 mol x (98.09g/1 mol)
= 14713.5g
=1.47 x 10^4

Is this right? The answer at the back of the textbook isn't the same, but my textbook has many errors. Is this one of them?

i really appreciate the help, thank you so much!
Answered by DrBob222
You were going great then you got de-railed.
Yes, mols = M x L
BUT C (that's molarity) = mols/L and that's by definition (and the 6.0 is in the problem) so sticking that 0.04 in there is a no, no.
mols = M x L = 6.0(from the problem) x 0.04 L = 0.24 mols
Then grams H2SO4 = mols H2SO4 x molar mass H2SO4 = 0.24 mols x 98 g/mol = 23.52g, then round to the correct number of significant figures. That 98 is an estimate on my part.
Answered by julie
Ohhh I get it, so I was supposed to change the formula to n= C x V? That makes a lot more sense, thank you so much!
Answered by DrBob222
Actually you didn't change the formula. The formula is M = mols/L and you just applied some algebra and rearranged it to mols = M x L. Note if you want L you would rearrange to L = mols/M
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