Suppose you were calibrating a 100.0 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 11°c instead. how is the mass of the 100.0 ml of distilled water you measured at 11°c different from the mass of 100.0 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 11°C and 20°C.
@20°C: Density = 0.99823 g/cm³ or g/mL @11°C: (10 - 11)/(10 - 20) = (0.99973 - Density)/(0.99973 - 0.99823) Solving for density: Density = 0.99958 g/cm³ or g/mL
Mass @ 20°C = 100 mL * 0.99823 g/mL = 99.823 g Mass @ 11°C = 100 mL * 0.99958 g/mL = 99.958 g Difference of Masses = |99.823 g - 99.958 g| = 0.135 g
