Question

$KF$ has $NaCl$ type structure. What is the distance between $K^{+}$ and $F^{-}$ in $KF$ if its density is $2.48$ g cm${}^{-3}$?

• A. $269$ pm
• B. $537.5$ pm
• C. $155.3\times {10}^{-24}$ cm
• D. $5.375$ cm

Answer 1

The correct option is A $269$ pm

The molar mass of KF is $58$ g/mol.

The edge length $=a=(\frac{n\times M}{d\times N_A}{)}^{\frac{1}{3}}$.

Substituting values in the above expression, we get

$a={\left(\frac{4\times 58}{2.48\times 6.023\times {10}^{23}}\right)}^{\frac{1}{3}}=5.375\times {10}^{-8}cm=537.5pm$

The distance between $K^{+}$ and $F^{-}$ ions is $\frac{a}{2}=\frac{537.8}{2}=268.8pm$