1.

Formula mass of `NaCl` is `58.45g mol^(-1)` and density of its pure form is `2.167g cm^(-3)`. The average distance between adjacent sodium and chloride ions in the crystal is `2.814xx10^(-8)cm`. Calculate Avogadro constant.

Answer» Density `=("Formula mass"xxz)/(Av. no . xxVolume )`
For `f.c.c.` structure of `NaCl` ,
`z=4(` for `f.c.c.), d=2.167g//cm^(3), ` formula mass
`=58.45g//mol,` edge length `=2(r^(+)+r^(-))`
`=2xx2.814xx10^(-8)=5.628xx10^(-8)`
`:.` Volume `=a^(3)=(5.628xx10^(-8))^(3)`
`=1.78xx10^(-22)cm^(3)`
`:. 2.167=(58.45xx4)/(Av. no . xx1.78xx10^(-22))`
`:. Av. no . =6.06xx10^(23)`


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