Question

At 25.0 °C the density of liquid water is 0.997 g/cm3, but at -10.0 °C the density of solid water (ice) is 0.917 g/cm3. If a 250.0 mL sample of liquid water originally at 25.0 °C is frozen and cooled to -10.0 °C, what volume will the solid occupy?

Answer 1

Answer:

The volume occupied by the solid water is 271 cubic centimeters or mililiters.

Explanation:

The water experiment a contraction, that is, a reduction in its occupied volume due to the change in temperature. But the mass remains constant according to Principle of Mass Conservation. Hence, we can predict the volume of water at -10.0 ºC by the following relationship:

$\rho_{A}\cdot V_{A} = \rho_{B}\cdot V_{B}$ (1)

$\rho_{A}$, $\rho_{B}$ - Densities of water at 25 ºC and - 10 ºC, measured in grams per cubic centimeter.

$V_{A}$, $V_{B}$ - Volume occupied by the water at 25 ºC and - 10 ºC, measured in cubic centimeters.

If $V_{A} = 250\,cm^{3}$, $\rho_{A} = 0.997\,\frac{g}{cm^{3}}$ and $\rho_{B} = 0.917\,\frac{g}{cm^{3}}$, then the volume occupied by the water at - 10 ºC is:

$V_{B} = \frac{\rho_{A}\times V_{A}}{\rho_{B}}$

$V_{B} = \frac{\left(0.997\,\frac{g}{cm^{3}} \right)\cdot (250\,cm^{3})}{0.917\,\frac{g}{cm^{3}} }$

$V_{B} = 271\,cm^{3}$

The volume occupied by the solid water is 271 cubic centimeters or mililiters.