# Question #4f7cf

##### 1 Answer

#### Explanation:

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, **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 **times denser** than water.

Since water's density is said to 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**.

**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)|)))#