Home
Class 12
CHEMISTRY
3.24 g of Hg(NO(3))(2) (molar mass = 324...

`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 :

A

`Hg(NO_(3))_(2)` and `HgCl_(2)` both are completely ionized .

B

`Hg(NO_(3))_(2)` is fully ionized but `HgCl_(2)` is fully unionized .

C

`Hg(NO_(3))_(2)` and `HgCl_(2)` both are completely unionized .

D

`Hg(NO_(3))_(2)` a is fully unionized but `HgCl_(2)` is fully ionized .

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 13|11 Videos

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 35|6 Videos

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 34|7 Videos

  • IONIC EQUILIBRIUM

    GRB PUBLICATION|Exercise All Questions|526 Videos

  • METALLUGY

    GRB PUBLICATION|Exercise Subjective type|1 Videos

Similar Questions

Explore conceptually related problems

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.

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 solution of sucrose (molar mass =342) is prepared by dissolving 688.4 g in 1000 g of water. Calculate The freezing point of solution.

0.01(M) solution an acid HA freezes at -0.0205^(@)C . If K_(f) for water is 1.86K kg mol^(-1) , the ionization constant of the conjugate base of the acid will be (consider molarity) ~= molarity)

20 g of a biaryelectrolyte (Molecular mass =100) are dissolved in 500 g of water. The freezing point of the solution is -0.74^(@)C and K_(f)=1.86 K kg mol^(-1) . The degree of ionisation of eletrolyte as :

What mass of ethylene glycol (molar mass = "62.0 g mol"^(-1) ) must be added to 5.50 kg of water to lower the freezing point of water from 0^(@)C to -10.0^(@)C ( k_(f) for water = "1.86 K kg mol"^(-1) ).

45 g of ethylene glycol (C_(2) H_(6)O_(2)) is mixed with 600 g of water. The freezing point of the solution is (K_(f) for water is 1.86 K kg mol^(-1) )

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