Question

Dichloroethane CH.CI (Given: density - 1.25 g/mL) Molecular Weight Volume of Sample (read meniscus) = 22.0 mL Mass of Sample (use given density and volume) = Moles in Sample = Molecules = Total Atoms =

Answer 1

duswer Dichloremethane Cahylla density density = 1.25g/m2 Molecular = (2x Mc) + ("XMA) + (2x Mu) weight (2712gmol) + + (4x! gmol') + ( 24 35.5 gmol) = (24 +4+71) gmolt 99gmolt Molecular weight = 99 99 gnuol" Volume of sample = 22.0 m2 Mass of sample = ? Mass = Volume x density 22.0 mL X 1.25 g/ml = 27.5g Mass = 27.5 g

Moles in sample = ? Mole Mass Molar mass 27.5 g li 0-2778 mol 99 gmolt Moles in sample= 0.2778 mol Molecules 2 No: 10.oly molecules = Moles y Avogadro number = 0.2778 X 6.022x183 molecules mal = 1.673 x 1023 molecules mal 1-67301023 molecules Molecules = in sample Total atoms =? Atoms in I molecule = = 8 Atoms in sample=8X1-673 x 103 atoms = 13.384 x 2023 atoms 13.384 x 1023 atoms No. of atoms