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The difference in height of the mercury column in two arms of U tube manometer in arrangement-I is h₁=660 mm . In another arrangement-II at same temperature, 222 gm of CaCl ₂ is dissolved in 324 gm of water and difference in height of mercury column in two arms is found to be h₂=680 ~mm . If the value of degree of dissociation for CaCl ₂ in arrangement-II is then the value of 6.4 is: [Atmospheric pressure =1 ~atm ]

Correct answer

4

Step-by-step solution

M _ CaCl ₂ =111 ~g n _ CaCl ₂ = 222 111 =2 mole n _ H ₂ O = 324 18 =18 ~mole Relative lowering in vapour pressure R.L.V.P. = P₀-P_S P₀ = n₁ i n₁ i+n₂ = 100-80 100 = i 2 i 2+18 or 0.2= 2 i 2 i+18 or 0.4 i+3.6=2 i i =2.25 So i =2.25 For CaCl ₂ i =1+( n -1) aligned & 2.25=1+(3-1) & = 1.25 2 =0.625 aligned 6.4 0.625=4 .

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