An experiment was performed in which an empty 100-mL graduatedcylinder was weigh

Question

An experiment was performed in which an empty 100-mL graduatedcylinder was weighed. It was weighed once again after it hadbeen filled to the 30.0 mL mark with dry sand. A 25 mL pipetwas used to transfer 25.00 mL of methanol to the cylinder. The sand-methanol was stirred until bubbles no longer emerged fromthe mixture and the sand looked uniformly wet. The cylinderwas then weighed again. Use the data obtained from theexperiment (and displayed at the bottom of this problem) to findthe density of the sand particles Mass of cylinder plus wetsand 88.6235 g Mass of cylinder plus drysand 68.8732 g Mass of emptycylinder   25.4608 g Volume of dry sand 30.0 mL Volume of sand + methanol 48.6 mL Volume of methanol 25.00 mL
An experiment was performed in which an empty 100-mL graduatedcylinder was weighed. It was weighed once again after it hadbeen filled to the 30.0 mL mark with dry sand. A 25 mL pipetwas used to transfer 25.00 mL of methanol to the cylinder. The sand-methanol was stirred until bubbles no longer emerged fromthe mixture and the sand looked uniformly wet. The cylinderwas then weighed again. Use the data obtained from theexperiment (and displayed at the bottom of this problem) to findthe density of the sand particles Mass of cylinder plus wetsand 88.6235 g Mass of cylinder plus drysand 68.8732 g Mass of emptycylinder   25.4608 g Volume of dry sand 30.0 mL Volume of sand + methanol 48.6 mL Volume of methanol 25.00 mL
Mass of cylinder plus wetsand 88.6235 g Mass of cylinder plus drysand 68.8732 g Mass of emptycylinder   25.4608 g Volume of dry sand 30.0 mL Volume of sand + methanol 48.6 mL Volume of methanol 25.00 mL Mass of cylinder plus wetsand 88.6235 g Mass of cylinder plus drysand 68.8732 g Mass of emptycylinder   25.4608 g Volume of dry sand 30.0 mL Volume of sand + methanol 48.6 mL Volume of methanol 25.00 mL

Explanation / Answer

        Mass of cylinderplus dry sand , m' = 68.8732 g          Mass of emptycylinder , m"= 25.4608 g         Volume of drysand , V'= 30.0 mL         Volume ofsand + methanol ,V = 48.6 mL         Volume ofmethanol, V" =25.00 mL Mass of dry sand = Mass of cylinder plus dry sand -Mass of empty cylinder                             =m' - m''                            = 43.4124 g Volume of dry sand , V'' = 30.0 mL Density of the dry sand = Mass of dry sand / Volumeof dry sand                                   = 43.4124 / 30                                   = 1.44708 g / mL
                            =m' - m''                            = 43.4124 g Volume of dry sand , V'' = 30.0 mL Density of the dry sand = Mass of dry sand / Volumeof dry sand                                   = 43.4124 / 30                                   = 1.44708 g / mL There are spaces between the sand particles and thefluid occupies these spaces. If I use that assumption tocalculate the density of sand particles, I'd have to say that thesand particles only occupy a volume of (48.6 -30.0 ) = 18.6mL. They weigh the same as the dryparticles, so use this mass to calculate the density of sandparticles.
However, if sand particles refers to sand+methanol, the usethe weight of the wet sand (88.6235 -25.4608 )g and volume 48.6mLto calculate the density of the sand particles.i.e.,density of thesand particles = 22.4296 / 48.6 = 1.299 g / mL

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