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)

A solid cube of edge length =25.32 mm of an ionic compound which has N

1.	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» Solution :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 forNaCl type unit cell, `a=2(r_(+)+r_(-)) " " IMPLIES " " r_(+)=100` pm