A solid cube of edge length =25.32 mm of an ionic compound which has NaCl type lattice is added to 1kg of water.The boiling point of this solution is found to be `100.52^@C`(assume 100% ionisation of ionic compound).If radius of anion of ionic solid is 200 pm then calculate radius of cation of solid in pm (picometer).
(`K_b` of water =0.52 K kg `"mole"^(-1)`,Avogadro's number,`N_A=6xx10^(23),(root3(75))=4.22`)

Answer

`r_+ =100`

Explanation

Effective molality of solution=1
Hence, number of moles of ionic solid in given cube=0.5
So, number of formula units in given cube `=0.5xx6xx10^(23)`
Number of units cells `=1/4xx0.5xx6xx10^23=7.5xx10^22`
number of unit cells along one edge of cube `=root(3)(75)xx10^7=4.22xx10^7`
edge length of unit cell =`a=(25.32xx10^(-3))/(4.22xx10^7)m=600xx10^(-12) m =600` pm
for NaCl type unit cell, `a=2(r_(+)+r_(-)) " " implies " " r_(+)=100` pm