The atomic fraction `(d)` of tin in bronze (fcc) with a density of `7717 kg m^(-3)` and a lattice parameter of `3.903 Å` is `(Aw Cu = 63.54, Sn = 118.7, 1 amu = 1.66 xx 10^(-27 kg))`

Options

A. 0.01
B. 0.05
C. 0.1
D. 3.8

Answer

B

Explanation

` d = ([sum ("number of atom of each kind") xx (M_w " of each kind ") xx 1.66 xx 10^(-27) kg ] )/(a^3)`
`7717 kg m^(-3) = ([("number of Sn atoms ") xx (118.7 xx 1.66 xx 10^(-27) ) + ("number of Cu atoms") xx (63.54 xx 1.66 xx 10^(-27))])/((3.903 xx 10^(-10))^3 m^3)`
`276.4 = n_(Sn) (118.7) + n_(Cu) (63.54)`
`4.35 = 1.86n_(Sn) + n_(Cu)`
`n_(Cu) = 4 rArr n_(Sn) = 0.188`
atomic fraction ` = (n_(Sn))/(n_(Sn) = n_(Cu)) =0.05`