What volume of #30%# #m/m# #H_3PO_4#, #rho_"acid"=1.020*g*cm^-3#, contains #0.2452*g# acid...?

1 Answer
anor277
Oct 28, 2017

Approx. #0.25*cm^3#....

Explanation:

We use the old relationship....

#"Molarity"="Moles of solute"/"Volume of solution"#

Here we require a molar quantity of....

#250xx10^-3*Lxx0.01*mol*L^-1=2.50xx10^-3*mol# with respect to #H_3PO_4#.......................................i.e. #2.50xx10^-3*molxx98.08*g*mol^-1=0.2452*g# with respect to #H_3PO_4#...

We are supplied with a #30%m/m# solution, for which #rho= 1.020*g*cm^-3#...and so we take the quotient...

#(0.2452*g)/(1.020*g*cm^-3)=0.240*cm^3#

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