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.
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!
3 answers
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!
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