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The molal freezing point for water is 1....

The molal freezing point for water is `1.86^(@)C mol^(-1)`. If 342g of cane sugar is dissolved in 1000 mL of water, the solution will freeze at

A

`1.86^(@)C `

B

`-1.86^(@)C `

C

`- 3.92^(@)C`

D

`3.92^(@)C `

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The correct Answer is:
B
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The molal freezing point constant of water is 1.86 K m^(-1) . If 342 g of cane sugar (C_(12)H_(22)O_(11)) is dissolved in 1000 g of water, the solution will freeze at

The molar freezing point constant for water is 1.86 ^@C / molal . If 342 gm of canesugar (C_(12)H_(22)O_(11)) are dissolved in 1000 gm of water, the solution will freeze at

The molal freezing point constant for water is 1.86^(@)C//m . Therefore, the freezing point of 0.1 M NaCl solution in water is expected to be:

3.24 g of Hg(NO_(3))_(2) (molar mass = 324) dissolved in 1000 g of water constitutes a solution having freezing point of -0.0558 ^(@)C while 21.68 g of HgCl_(2) (molar mass = 271) in 2000 g of water constitutes a solution with a freezing point of 0.0744^(@)C . The K_(f) for water is 1.86 (K-Kg)/("Mol") . About the state of ionization of these two solids in water it can be inferred that :

3.24gHg(NO_(3))_(2) (molecules mass 324 )dissolved in 1000 g of water constitutes a solution having a freezing point -0.0558^(@)C while 21.68g of HgCl_(2) (Molecular mass 271 in 2000g of water constitutes a solution with a freezing point of -0.0744^(@)C .The K_(f) of water is 1.86K Kg mol^(1) .About the state of ionisation of the two solids in water can be inferred that.

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