calculate the value of Na(avagadro no.) from following data :- KF = 2.48 g/cm³ .Distance between K+ nd F- in KF is 269 pm .Atomic mass of K is 39 nd F is 19 amu ........no spam please give satisfying answer...don't google past nd screenshot frm google etc.....thank u
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here is ur answer
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Distance between K+ and F- = 269 pm
Edge length (a) = 2 x 269 = 538 pm.
= 538 x 10-10 cm
Mass of K+ F- unit = 39+19=58=(M)
Density of KF = 2.48 g cm-3
(z) no of K+ F- units in KF crystals = 4k+ F- = 4
N0 = ?
r Z x M / a3 x N0 N0 = Z x M / a3 x r
= 4 x 58 / (538 x 10-10)3 x 2.48
= 6.02 x 1023 atoms mol-1
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cya
here is ur answer
=======
Distance between K+ and F- = 269 pm
Edge length (a) = 2 x 269 = 538 pm.
= 538 x 10-10 cm
Mass of K+ F- unit = 39+19=58=(M)
Density of KF = 2.48 g cm-3
(z) no of K+ F- units in KF crystals = 4k+ F- = 4
N0 = ?
r Z x M / a3 x N0 N0 = Z x M / a3 x r
= 4 x 58 / (538 x 10-10)3 x 2.48
= 6.02 x 1023 atoms mol-1
=====
cya
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