1.

KF crystallizes in fce structure like sodium chloride. calculate the distance between K^(+) and F^(-) in KF. ("given : density of KF is 2.48 g cm"^(-3))

Answer»

Solution :`"DENSITY of KF"="2.48 g cm"^(-3)`
`RHO=(nM)/(a^(3)N_(A))`
N = 4 (For Fcc)
`rho=(4xx58)/(a^(3)xx6.023xx10^(23))`
`2.48=(4xx58)/(a^(3)xx6.023xx10^(23))`
`a^(3)=(4xx58)/(2.48xx6.023xx10^(23))`
`a^(3)=(232)/(14.93xx10^(23))`
`a^(3)=15.54xx10^(-23)`
`a="537.5 pm"`
`d=(a)/(sqrt2)"(For fcc)"`
`d=(537.5)/(1.414)`
`d=380.12" pm"`


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