For an ionic solid `MX_(2)`, where X is monovalent, the enthalpy of formation of the enthalpy of formation of the solid from M (s) and `X_(2) (g)` is 1.5 times the electron gain enthalpy of `X(g)`. The first and second ionisation enthalpies of the meta (M) are 1.2 and 2.8 times of the enthalpy of sublimation of `M(s)`. The bond dissociation enthalpy of `X_(2) (g)` is 0.8 times the first ionisation enthalpy of metal and it is also equal to one -fifth of the magnitude of lattice enthalpy of `MX_(2)`. If the electron gain enthalpy of `X (g)` is `-96 K cal//mol`, the answer the enthalpy of sublimation of metal (M) in K cal/mol

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Correct Answer - 41.38
`{:(M(s) rarr M(g)" "DeltaH=x),(M(g) rarr M^(+)(g)" "DeltaH=1.2 x),(M^(+)(g) rarrM^(2+)(g)" "DeltaH=2.8 x),(X_(2) rarr 2X(g)" "DeltaH=0.8xx1.2 x=1/5 z),(2X(g) rarr 2X^(-) (g)" "DeltaH=-2y),(M^(2+)(g)+2X^(-) (g) rarr MX_(2)(s)" "DeltaH=-z=-4.8x),(bar(M(s)+X_(2)(g) rarr MX_(2)(s)" "DeltaH=-1.5 y" ")):}`
`x+4x+0.96x-2y-4.8 x=-1.5 y`
`1.16x=+0.5 y" "rArrx=(0.5xx96)/(1.16)=41.38`

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