University of Kentucky-CHE × E Chemistry of Pt ng.com/ibiscms/modfibis/view.php?

Question

University of Kentucky-CHE × E Chemistry of Pt ng.com/ibiscms/modfibis/view.php?id-4865409 Kentucky-CHE 107-Spring18-DEPTActivities and Due Dates HW #2 1/25/2018 11:59 PM 3 90/100 Gradebo Print CalcuiatorPeriodic Table Question 12 of 15 IncorrectIcorect Sapling Learning A metal crystallizes in the face-centered cubic (FCC) lattice. The density of the metal is 8960 kg/m3 and the length of a unit cell edge, a, is 361.47 pm. Calculate the mass of one metal atom Number Tools x 10 Identify the metal O nickel O silver O rhodiunm copper O Prevous Give Up & View Sollion 9 Check AnswerNext Eil , Hint

Explanation / Answer

given

edge length (a) = 361.47 pm = 361.47 x 10-12 m

now

volume of unit cell = a^3 = (361.47 x 10-12 m )^3

volume of unit cell = 4.723 x 10-29 m3

now

mass of metal in unit cell = volume of unit cell x density

mass of metal in unit cell = 4.723 x 10-29 m3 x 8960 kg/m3

mass of metal in unit cell = 4.2318 x 10-25 kg

now

there will be 4 metal atoms per unit cell in FCC

so

mass of one metal atom = (4.2318 x 10-25 ) / 4 = 1.058 x 10-25 kg

mass of one metal atom = 1.058 x 10-25 kg x 1000 g / 1kg

mass of one metal atom = 1.058 x 10-22 g

b)

now

atomic mass of metal = mass of one atom x avagadro number

atomic mass of metal = mass of one atom x 6.022 x 10^23 atoms / mol

atomic mass of metal = (1.058 x 10-22 g / atom) x (6.022 x 10^23 atoms / mol)

atomic mass of metal = 63.71 g /mol

so

the metal is copper

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