3.24 g Hg(NO3)2 (Molecular mass 324) dissolved in 1000 g of water constitutes a solution having a freezing point - 0.0558°C, while 21.68 g of HgCl2 (Molecular mass 271) in 2000 g of water constitutes a solution with a Freezing Point of - 0.0744°C. The Kf of water is 1.86 Kg mol-1 About the state of ionisation of the two solids in water can be inferred that

Question

3.24 g Hg(NO3)2 (Molecular mass 324) dissolved in 1000 g of water constitutes a solution having a freezing point - 0.0558°C, while 21.68 g of HgCl2 (Molecular mass 271) in 2000 g of water constitutes a solution with a Freezing Point of - 0.0744°C. The Kf of water is 1.86 Kg mol-1 About the state of ionisation of the two solids in water can be inferred that (A) Hg(NO3)2 is fully ionised but HgCl2 is fully unionised (B) Hg(NO3)2 and HgCl2 both are completely ionised (C) Hg(NO3)2 is fully unionised but HgCl2 is fully ionised (D) Both Hg(NO3)2 and HgCl2 are completely unionised

Tardigrade Admin · Accepted Answer

0.0558 = i × (3.24/324) × 1.86 i=3 (100% ionised) 0.0744 = i × (21.68/271) × 1.86 i=1 (unionised)

$H g (N O_{3})_{2}$ is fully ionised but $H g C l_{2}$ is fully unionised

$H g (N O_{3})_{2}$ and $H g C l_{2}$ both are completely ionised

$H g (N O_{3})_{2}$ is fully unionised but $H g C l_{2}$ is fully ionised

Both $H g (N O_{3})_{2}$ and $H g C l_{2}$ are completely unionised

$0.0558 = i \times \frac{3.24}{324} \times 1.86$
$i = 3$ (100% ionised)
$0.0744 = i \times \frac{21.68}{271} \times 1.86$
$i = 1$ (unionised)

Hg(NO3​)2​ is fully ionised but HgCl2​ is fully unionised

Hg(NO3​)2​ and HgCl2​ both are completely ionised

Hg(NO3​)2​ is fully unionised but HgCl2​ is fully ionised

Both Hg(NO3​)2​ and HgCl2​ are completely unionised

0.0558=i×3243.24​×1.86 i=3 (100% ionised) 0.0744=i×27121.68​×1.86 i=1 (unionised)

$H g (N O_{3})_{2}$ is fully ionised but $H g C l_{2}$ is fully unionised

$H g (N O_{3})_{2}$ and $H g C l_{2}$ both are completely ionised

$H g (N O_{3})_{2}$ is fully unionised but $H g C l_{2}$ is fully ionised

Both $H g (N O_{3})_{2}$ and $H g C l_{2}$ are completely unionised

$0.0558 = i \times \frac{3.24}{324} \times 1.86$
$i = 3$ (100% ionised)
$0.0744 = i \times \frac{21.68}{271} \times 1.86$
$i = 1$ (unionised)