Question #4f7cf

Here's an intuitive way of approaching this problem.

Use the known mass of the flask to determine the mass of water it contained when filled with water

#m_"water" = "593.63 g" - "78.23 g" = "515.4 g H"_2"O"#

Now use the mass of the flask to determine the mass of sulfuric acid, #"H"_2"SO"_4#, it contained when filled with sulfuric acid

#m_("H"_2"SO"_4) = "1026.57 g" - "78.23 g" = "948.34 g H"_2"SO"_4#

Now, you don't need to know the actual volume of the flask. All you need to know is that the same volume contained

#(948.34 color(red)(cancel(color(black)("g"))))/(515.4color(red)(cancel(color(black)("g")))) = 1.84#

times more mass of sulfuric acid than of water. This means that the sulfuric acid solution is #1.84# times denser than water.

Since water's density is said to be #"1.00 g cm"^(-3)#, you can say that the density of the sulfuric acid solution will be

#rho_("H"_2"SO"_4) = 1.84 xx "1.00 g cm"^(-3) = color(green)(|bar(ul(color(white)(a/a)color(black)("1.84 g cm"^(-3))color(white)(a/a)|)))#

The answer is rounded to three sig figs.

#color(white)()#
ALTERNATIVE APPROACH

A quick way to double-check the answer is to actually calculate the volume of the flask. Since you know the mass of water it held when filled and the density of water, you can say that its volume is

#515.4 color(red)(cancel(color(black)("g"))) * overbrace("1 cm"^(-3)/(1.00color(red)(cancel(color(black)("g")))))^(color(blue)("density of water")) = "515.4 cm"^(-3)#

Now that you know the mass of sulfuric acid solution and the volume it occupies, you can find its density

#rho_("H"_2"SO"_4) = "948.34 g"/"515.4 cm"^(-3) = color(green)(|bar(ul(color(white)(a/a)color(black)("1.84 g cm"^(-3))color(white)(a/a)|)))#