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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

`2.42^(@)C`

D

`-2.42^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
B

`Delta T_(f) = K_(b) xx W_(2)/(M_(2) xx W_(1) (kg))`
`Delta T_(f) = 1.86 xx 342/(342 xx 1 ) = 1.86^(@)C`
(for water d=1g ` "mol"^(-1))`
`T_(f) = T^(@)- Delta T_(f) = 0^(@)C-1.86^(@)C=-1.86^(@)C`
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Knowledge Check

  • 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

    A
    `-1.86^(@)C`
    B
    `1.86^(@)`
    C
    `-3.92^(@)C`
    D
    `2.42^(@)C`
  • 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

    A
    `-1.86 ^@C`
    B
    `1.86^@C`
    C
    `-3.92^@C`
    D
    `2.42^@C`
  • The molal freezing point constant for water is 1.86^(@)C//m . Therefore, the freezing point of 0.1 M

    A
    `-272.628K`
    B
    `+272.628K`
    C
    `-0.372^(@)C`
    D
    `+0.372^(@)C`
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    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 :

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