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A patient turns in a sample of urine for specific gravity test. The lab results record a value of1.024. What is the density of this patient's urine?

User Peshal
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Final answer:

The density of the patient's urine is 1.024 g/mL, as specific gravity is the ratio of a substance's density to that of pure water, which has a density of 1.00 g/mL. Variations in specific gravity reflect changes in solute concentration and can indicate different medical conditions.

Step-by-step explanation:

The lab results recorded a specific gravity of 1.024 for a patient's urine sample. Specific gravity is the ratio of the density of a substance to the density of pure water under the same conditions. Since the density of pure water is 1.00 g/mL, the patient's urine will have a density greater than that of pure water. Therefore, with a specific gravity of 1.024, the density of the urine is 1.024 g/mL.

An increase in the specific gravity of urine indicates higher solute concentration, which could be due to various conditions such as dehydration, diarrhea, or infection. Conversely, a decrease in solute concentration, resulting in lower specific gravity, could indicate medical concerns like renal failure. Monitoring the specific gravity of urine is important in medical diagnostics and can assist in the discovery of various health conditions.

User Machaku
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Specific gravity is the same concept for relative density. This measure indicates the density with respect to the density of water at standard temperature. The formula is:


\text{Specific gravity = }\frac{density\text{ of substance}}{density\text{ of water}},

If we are comparing pure water, the density of this would be 1.00 g/mL, so clearing the formula and replacing, we're going to have:


\begin{gathered} \text{density of substance (urine) = specific gravity }\cdot\text{ density of water,} \\ \text{density of urine = 1.024 }\cdot\text{ 1.00}(g)/(mL), \\ \text{density of urine = 1.024 }(g)/(mL). \end{gathered}

The density of the patient's urine is 1.024 g/mL if the specific gravity is 1.024.

User PramodValavala
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