Home
Class 12
CHEMISTRY
The freezing point depression of 0.001 m...

The freezing point depression of `0.001 m K_(x)` `[Fe(CN)_(6)]` is `7.10xx10^(-3) K`. Determine the value of x. Given, `K_(f)=1.86 K kg "mol"^(-1)` for water.

Text Solution

Verified by Experts

`Deltax = ixx K_(F) xx m`
`7.10 xx 10^(-3) =ixx 1.86 xx 0.001`
`i= 3.817`
`alpha = (i-1)/(n-1)`
`1 = (3.817-1)/((x+1)-1)`
`x = 2.817 ~~ 3`
`:.` Molecular formula of the compound is `K_(3)[Fe(CN)_(6)]`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    ALLEN|Exercise EXERCISE -01|53 Videos

  • SOLUTIONS

    ALLEN|Exercise EXERCISE -02|44 Videos

  • SOLUTIONS

    ALLEN|Exercise EXERCISE -05 [B]|22 Videos

  • S-BLOCK ELEMENTS

    ALLEN|Exercise EXERCISE -3|1 Videos

  • Some Basic Concepts of Chemistry (Mole concept)

    ALLEN|Exercise All Questions|39 Videos

Similar Questions

Explore conceptually related problems

Freezing point depression of millimolal K_nFe(CN)_(6) is 7.1 xx 10^(-3) K. If K_f = 1.86 k Kg mol^(-1) . Calculate the value of n.

The freezing point (.^(@)C) of a solution containing 0.1 g of K_(3)[Fe(CN)_(6)] (molecular weight 329) on 100 g of water (K_(f)=1.86 K kg mol^(-1))

The freezing point of a 1.00 m aqueous solution of HF is found to be -1.91^(@)C . The freezing point constant of water, K_(f) , is 1.86 K kg mol^(-1) . The percentage dissociation of HF at this concentration is:-

45 g of ethylene glycol C_(2)H_(6)O_(2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of solution. Given K_(f)=1.86 K kg mol^(-1) .

45 g of ethylene glycol C_(2)H_(4)O_(2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of solution. Given K_(f)=1.86 K kg mol^(-1) .

0.01 m aqueous solution of K_(3)[Fe(CN)_(6)] freezes at -0.062^(@)C . What is the apparent percentage of dissociation ? ( K_(f) for water = 1.86" K kg mol"^(-1) )

The freezing point of a solution containing 0.1g of K_3[Fe(CN)_6] (Mol. Wt. 329) in 100 g of water ( K_f =1.86K kg mol^(−1) ) is:

0.01 molal aqueous solution of K_(3)[Fe(CN)_(6)] freezes at -0.062^(@)C . Calculate percentage dissociation (k_(f)=1.86)

The freezing point depression of 0.1 molal NaCl solution is 0.372 K. What conclusion can you draw about the molecular state of NaCl in water. K_f of water = 1.86 k/m.

To 250 mL of water, x g of acetic acid is added. If 11.5% of acetic acid is dissociated, the depressin in freezing point comes out 0.416 . What will be the value of x if K_(f) ("water")=1.86 K kg^(-1) and density of water is 0.997 g mL .