Answer
See explanation
Work Step by Step
The gas bubbles are air in the sand escaping; methanol and sand are not reacting. We will assume that the mass of trapped air is insignificant.
Mass of dry sand = 37.3488 g − 22.8317 g = 14.5171 g
Mass of methanol = 45.2613 g − 37.3488 g = 7.9125 g
The volume of sand particles (air absent) = volume of sand and methanol − the volume of methanol
Volume of sand particles (air absent) = 17.6 mL − 10.00 mL = 7.6 mL
The density of dry sand (air present) = 14.5171g/ 10.0 mL = 1.45 g/mL
Density of methanol = 7.9125 g/10.00 mL = 0.7913 g/mL
Density of sand particles (air absent) = 14.5171g/7.6 mL = 1.9 g/mL
