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Q.

When one mole of A is reacted completely in an ice calorimeter at 0° C and 1 atm, it is found that the volume of equilibrium mixture of ice and water decreased by 0.25 mL. The .ΔrH273 for the reaction
2AB+2C
is: (the densities of water and ice at 0° and 1 atm are 1.00 and 0.96 gm/cm3 , respectively and latent heat of fusion of ice at 0° C and 1 atm is 80 cal/gm)
 [Use : 1 if .ΔrH273 is +ve and 2 if .ΔrH273 is -ve and answer as 1 abc or 2 abc, where abc is magnitude of ΔrH273]

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answer is 2959.

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Detailed Solution

H2O(s)H2O(l);ΔH1=+ ve 
Volume decrease hence reaction shifts in forward direction 
2AB+2C 

ΔrH273K=ve
Change in volume of H2O(s)H2O(l)
at 1 atm and 273 k
0.25mL=Vice VH2O(h)=xmol180.96181xmol=13mol 
Hence, energy gain by ice
ΔrH273=80×13×18
                  = 480 cal from one mol of A

2AB+2C; ΔrH=2×480=960cal=2960

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