Question

A 50-mL pycnometer is found to weigh 120 g when empty, 171 g when filled with water, and 160 g when filled with an unknown liquid. Calculate the specific gravity of the unknown liquid.

Answer 1

We know that the specific gravity is a relation between densities:

$\text{Specific gravity}=\cfrac{\text{density of unknown liquid}}{\text{density of water}}$

We proceed to use the definition of density to find the specific gravity:

$\text{Specific gravity}=\cfrac{\cfrac{m_{\text{unknown liquid}}}{V_{pycnometer}}}{\cfrac{m_{\text{water}}}{V_{pycnometer}}}=\cfrac{m_{\text{unknown liquid}}}{m_{water}}$

We proceed to calculate the result with the mass of the unknown liquid and water as:

$\text{Specific gravity}=\cfrac{m_{\text{unknown liquid}}}{m_{water}}=\cfrac{(160-120)g}{(171-120)g} \\ \text{ } \\ \implies \text{Specific gravity}=40/51=0.7843$

In conclusion, the specific gravity of the unknown liquid is 0.7843.

Reference

• Chang, R., & Goldsby, K. A. (2010). Chemistry. Chemistry, 10th ed.; McGraw-Hill Education: New York, NY, USA.