Question

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

Answer 1

`1.01`

Explanation:

The specific gravity of a substance, `"SG"`, is simply the ratio between that substance's density and the density of a reference substance, which is usually water at `4^@"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

`"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 `"1.014 g mL"^(-1)`, i.e. every milliliter of urine has a mass of `"1.014 g"`.

Now, water has a maximum density of `"0.999975 g mL"^(-1)` at `4^@"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.