Chapter 1, Problem 1.141QP

Interpretation Introduction

Interpretation:

The specific gravity of a dehydrated patient urine sample is $\text{1}\text{.04}$.  Using the density of water at ${\text{37}}^{\text{°}}\text{C,}\text{0}\text{.993}\text{g/mL}$, the density of the urine sample has to be calculated.

Concept Introduction:

Specific gravity:

The reference of a value of density to the density of pure water, at ${\text{4}}^{\text{°}}\text{C}$ is said to be the specific gravity.  It is a unit less term.

  $\text{specific}\text{gravity}\text{=}\frac{\text{density}\text{of}\text{object}\text{(g/mL)}}{\text{density}\text{of}\text{water}\text{(g/mL)}}$