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

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Answer 1 · 2016-05-26T06:03:57

Under $1.5*mL$ .

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

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?

What volume of 18.0 M H_2SO_4H_2SO_4 is needed to contain 2.45 g H_2SO_4H_2SO_4?