Find the freezing point of a solution when 5 mol of KI is mixed with 1 mol of HgCl2 in 3500 gram of H2OKf of H2O=1.86Km⋅KI+HgCl2→K2Hgl4+KCl

4KI+HgCl2→K2Hgll4+2KClRemained KI=5-4=1 mol1 mol K2Hgl4 and 2 mol KCl is formed K2Hgl4→2K++Hgll42−ΔTf0=Kfi1m1+i2m2+i3m3Tf0=−4.780C

Find the freezing point of a solution when 5 mol of KI is mixed with 1 mol of HgCl2 in 3500 gram of H2OKf of H2O=1.86Km⋅KI+HgCl2→K2Hgl4+KCl

4KI+HgCl2→K2Hgll4+2KClRemained KI=5-4=1 mol1 mol K2Hgl4 and 2 mol KCl is formed K2Hgl4→2K++Hgll42−ΔTf0=Kfi1m1+i2m2+i3m3Tf0=−4.780C

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Find the freezing point of a solution when 5 mol of KI is mixed with 1 mol of HgCl₂ in 3500 gram of H₂O
$K_{f} of H_{2} O = 1 .86 \frac{K}{m} \cdot (KI + {HgCl}_{2} \to K_{2} {Hgl}_{4} + KCl)$

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$- 28 {.7}^{0} C$

$- {41.45}^{0} C$

$- 35 {.07}^{0} C$

$- 4 {.78}^{0} C$

answer is D.

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Detailed Solution

$4 KI + {HgCl}_{2} \to K_{2} {Hgll}_{4} + 2 KCl$
Remained KI=5-4=1 mol
1 mol K₂Hgl₄ and 2 mol KCl is formed $(K_{2} {Hgl}_{4} \to 2 K^{+} + {Hgll}_{4}^{2 -})$
$\begin{array}{l} |{({ΔT}_{f})}_{0}| = K_{f} (i_{1} m_{1} + i_{2} m_{2} + i_{3} m_{3}) \\ {(T_{f})}_{0} = - 4 {.78}^{0} C \end{array}$