What volume of #84%# #HNO_3# #"w/w"#, #rho=1.54*g*mL^-1#, is required to make a #1.0*L# of a #1*mol*L^-1# solution of nitric acid?

1 Answer
Sep 3, 2017

Answer:

Approx. #25*mL#

Explanation:

We use the relationship, #"concentration"="moles of solute"/"volume of solution"#

And thus we need #1*Lxx0.50*mol*L^-1=0.50*mol# with respect to nitric acid. And this represents a mass of #0.50*molxx63.01*g*mol^-1=31.5*g#.

We gots #84%# #HNO_3#, #"w/w"#............

And thus we need a mass of solution......

#(31.5*g)/(0.84)=37.5*g#.......

And given the density......#rho=1.54*g*mL^-1#, we need....

#(37.5*g)/(1.54*g*mL^-1)=24.4*mL.#

Nitric acid of this concentration is called in the trade #"red fuming nitric acid"#, and its density owes to the solubility of #NO_2#.