What volume of 18.0 M #H_2SO_4# is needed to contain 2.45 g #H_2SO_4#?

1 Answer
May 26, 2016

Answer:

Under #1.5*mL#.

Explanation:

#(2.45*g)/(98.08*g*mol^-1)# #=# #0.0250*mol" sulfuric acid"#.

We have #18.0*mol*L^-1# #H_2SO_4# available.

Volume required #=# #(0.0250*mol)/(18.0*mol*L^-1)xx1000*mL*L^-1# #=# #??*mL#

PS Have you ever lifted a #2.5*L# bottle of conc. sulfuric? Why is it so heavy?