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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?

Respuesta :

meerkat18
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
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