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 had beenfilled to the 10.0-mL mark with dry sand. A 10-mL pipet was used totransfer 10.00-mL of methanol to the cylinder. The sand-methanolmixture was stirred until bubbles no longer emerged from themixture and the sand looked uniformly wet. The cylinder was thenweighed again. Use the data obtained from this experiment(anddisplayed at the end of this problem) to find the density of thedry sand, the density of methanol, and the density of sandparticles. Does the bubbling that occurs when the methanol is addedto the dry sand indicate that the sand and methanol are reacting? Mass of cylinder plus wetsand        45.2613 g Mass of cylinder plus drysand         37.3488g Mass of emptycylinder                    22.8317g Volume of drysand                         10.0mL Volume of sand +methanol              17.6mL Volume ofmethanol                        10.00mL Mass of cylinder plus wetsand        45.2613 g Mass of cylinder plus drysand         37.3488g Mass of emptycylinder                    22.8317g Volume of drysand                         10.0mL Volume of sand +methanol              17.6mL Volume ofmethanol                        10.00mL

Explanation / Answer

         Mass ofcylinder plus dry sand , m' = 37.3488 g          Mass of emptycylinder , m"= 22.8317 g         Volume of drysand , V'= 10.0 mL         Volume ofsand + methanol ,V = 17.6 mL         Volume ofmethanol, V" =10.00mL Mass of dry sand = Mass of cylinder plus drysand - Mass of empty cylinder                             =m' - m''                            = 14.5171 g Volume of dry sand , V'' = 10.0 mL Density of the dry sand = Mass of dry sand / Volumeof dry sand                                   = 14.5171 /10                                   = 1.45171 g / mL Mass of methanol = Mass of cylinder plus wet sand - Mass ofcylinder plus dry sand                            = m - m'                            = 45.2613 - 37.3488                            = 7.9125 g Volume of methnol = 10.0 mL Density of methanol = mass of methanol / Volume ofmethanol                                      =7.9125 / 10                                     = 0.79125 g / mL There are spaces between the sand particles and the fluidoccupies these spaces. If I use that assumption to calculatethe density of sand particles, I'd have to say that the sandparticles only occupy a volume of (17.6-10.0 ) = 7.6mL.They weigh the same as the dry particles, so use this mass tocalculate the density of sand particles

However, if sand particles refers to sand+methanol, the use theweight of the wet sand (45.2613-22.8317)g and volume 17.6mL tocalculate the density of the sand particles.i.e.,density of thesand particles = 22.4296 / 17.6 = 1.2744 g / mL

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.