# A 50.0 mL urine sample has a mass of 50.7 g. What is the specific gravity of the urine?

##### 1 Answer

#### Answer:

#### Explanation:

The **specific gravity** of a substance, *density* and the density of a reference substance, which is usually water at

So right from the start, the fact that specific gravity is a **ratio** of two densities should let you know that you're looking for a *unitless quantity*.

In your case, the specific gravity of urine will be

#"SG"_ "urine" = rho_"urine"/rho_"water"#

Use the mass and volume of the sample to calculate its density, which is essentially the mass of **one unit of volume* of a given substance

#1 color(red)(cancel(color(black)("mL"))) * "50.7 g"/(50.0color(red)(cancel(color(Black)("mL")))) = "1.014 g"#

This means that urine has a density of **every milliliter** of urine has a mass of

Now, water has a maximum density of

#"SG"_ "urine" =(1.014 color(red)(cancel(color(black)("g mL"^(-1)))))/(0.999975color(red)(cancel(color(black)("g mL"^(-1))))) = color(green)(bar(ul(|color(white)(a/a)color(black)(1.01)color(white)(a/a)|))) #

The answer is rounded to three **sig figs**.