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

Nov 8, 2016

$1.01$

#### Explanation:

The specific gravity of a substance, $\text{SG}$, is simply the ratio between that substance's density and the density of a reference substance, which is usually water at ${4}^{\circ} \text{C}$, the temperature at which its density is maximum.

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

$\text{SG"_ "urine" = rho_"urine"/rho_"water}$

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 ${\text{1.014 g mL}}^{- 1}$, i.e. every milliliter of urine has a mass of $\text{1.014 g}$.

Now, water has a maximum density of ${\text{0.999975 g mL}}^{- 1}$ at ${4}^{\circ} \text{C}$, so use this value to find the specific gravity of urine

"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.