Question

The molarity of a 80.3% ( by weight ) H2SO4 solution is x , and it contains 91g solvent per 100 ml of solution . The value of x is :

Answer 1

In 100 mL of solution 91 g are solvent this is 19.7% (by mass) of the solution

We make a ratio to find the mass of \(H_2SO_4\)

\(\sf \dfrac{19.7\%}{80.3\%}=\dfrac{91~g}{m_{H_2SO_4}}\rightarrow m_{H_2SO_4}=37.09~g\)


Use Molarity formula:

\(\sf Molarity=\dfrac{n_{solute}}{L_{solution}}=\dfrac{\dfrac{m_{H_2SO_4}}{M_{H_2SO_4}}}{L_{solution}}=\dfrac{\dfrac{37.09~g}{98.08~g/mol}}{0.1~L}=3.78190~M\approx 3.8~M\)

Answer 2

Answer is 9 :(

Answer 3

also there was a small error , mH2SO4 = 370.9 , but still answer does not match :(

Answer 4

crap, yeah, i think i used 9.1 g not 91 g.

If the answer is 9 M, then you'd have to have

\(\sf 9~M=\dfrac{m_{H_2SO_4}}{(0.1~L*98.08~g/mol)}\rightarrow m_{H_2SO_4}=88.272~g\)

that's not possible if the solution is 80.3% ( by weight ) H2SO4.

I think there might be an error in the question here "91g solvent per 100 ml of solution", because the density of water (which is probably the solvent) is 1 g/mL, so that would compose ~91 mL of solution which would not be 19.7% of the mass.

Answer 5

Okay , Thanks a ton ! :)