Suppose you were calibrating a 50.0 ml volumetric flask using distilled water. the flask temperature was at 20°c, and you assumed that the distilled water was as well. however, you later discover that the actual water temperature was 14°c instead. how is the mass of the 50.0 ml of distilled water you measured at 14°c different from the mass of 50.0 ml of distilled water at 20°c?

Density is sensitive to temperature for gases and liquids, although not much for liquids. We use the data in the picture. Using linear interpolation, we determine the densities at 14°C and 20°C.

@20°C: Density = 0.99823 g/cm³ or g/mL
@14°C:
(10 - 14)/(10 - 20) = (0.99973 - Density)/(0.99973 - 0.99823)
Solving for density:
Density = 0.99913 g/cm³ or g/mL

Mass @ 20°C = 50 mL *  0.99823 g/mL = 49.9115 g
Mass @ 14°C = 50 mL *  0.99913 g/mL = 49.9565 g
Difference of Masses = |49.9115 g - 49.9565 g| = 0.045 g

The correct answer is b