One soluble 18.0% ammonium sulfate has a density of 1.10 g / mL. a) How many grams (NH4)2SO4 are needed to prepare 275 mL of this solution? b)How much mL of this solution are found im 90.0 g of ammonium sulfate?

1 Answer
Feb 16, 2018

So you gots #18%*m/m# #"ammonium sulfate"# with #rho_"solution"=1.10*g*mL^-1#...we need to dissolve approx. #50*g#...

Explanation:

We work out (i), the concentration of this solution in #mol*L^-1#:

...we take a #1*mL# volume whose MASS is #1.10*g#, 18% of which is due to the salt....

#[(NH_4)_2SO_4]=((18%xx1.10*g)/(132.14 *g*mol^-1))/(1.00xx10^-3*L)=1.50*mol*L^-1#

And (ii), we work out the mass of solute required for the specified VOLUME....

We know that #"concentration"="moles"/"volume"#

And thus #"volume"xx"concentration"="moles"#

#275*mLxx10^-3*L*mL^-1xx1.50*mol*L^-1=0.412*mol#

And (iii), this represents a mass of #0.412*molxx132.14 *g*mol^-1=??*g#

And (iv), if we got a volume of this solution THAT contains #90*g# of salt....this represents a molar quantity of #(90*g)/(132.14*g*mol^-1)=0.681*mol#...

And to get the required volume we take the quotient...

#(0.681*mol)/(1.50*mol*L^-1)xx10^3*mL*L^-1=454.0*mL#

And you can check the arithmetic. All care taken, but no responsibility admitted....