Home
Class 12
CHEMISTRY
The molal depreesion constant of water i...

The molal depreesion constant of water is `1.86^(@)C//m`. Then find out the freezing point of 0.1 m NaCl solution.

A

`-1.86^(@)C`

B

`0.372^(@)C`

C

`-0.372^(@)C`

D

`-0.186^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
C
For NaCl, i = 2
`Delta T_(f)=2K_(f).m`
`= 2xx1.86xx0.1=0.372`
`T_(s)=T-Delta T_(f)=0-0.372=-0.372^(@)C`.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • SOLUTIONS

    KUMAR PRAKASHAN|Exercise SECTION - E MCQs asked in JEE/NEET/AIEEE|41 Videos

  • SOLUTIONS

    KUMAR PRAKASHAN|Exercise SECTION - E MCQs asked in GUJCET/Board Exam|69 Videos

  • SOLUTIONS

    KUMAR PRAKASHAN|Exercise SECTION - E Assertion and Reason Type|5 Videos

  • SAMPLE QUESTION PAPER

    KUMAR PRAKASHAN|Exercise [PART-B] SECTION - C|2 Videos

  • SURFACE CHEMISTRY

    KUMAR PRAKASHAN|Exercise SECTION - E (MCQs asked in GUJCET/Board Exams)|52 Videos

Similar Questions

Explore conceptually related problems

The molal depression constant for water is 1.86^(@)C . The freezing point of a 0.05-molal solution of a non-electrolyte in water is

The boiling point of an aqueous solution of a non-volatile solute is 100.15^(@)C . What is the freezing point of an aqueous solution obtained by dilute the above solution with an equal volume of water. The values of K_(b) and K_(f) for water are 0.512 and 1.86^(@)C mol^(-1) :

Knowledge Check

  • The depression in freezing point for 0.01 m aqueous solution of Kx[Fe(CN)6] is 0.0744 K. The molal depression constant for solvent is 1.86 K kg mol–1. If the solute undergoes complete dissociation, what is the correct molecular formula for the solute?

    A
    `K_2[Fe(CN)_6]`
    B
    `K_3[Fe(CN)_6]`
    C
    `K[Fe(CN)_6]`
    D
    `K_4[Fe(CN)_6]`

  • The freezing point of 0.01 M KCl solution is -2^(@)C . If BaCl_(2) is completely ionized, what is the freezing point of 0.01 M BaCl_(2) solution ?

    A
    `-3^(@)C`
    B
    `+3^(@)C`
    C
    `-2^(@)C`
    D
    `-4^(@)C`

  • The degree of ionization of 0.2 m weak acid HX is 0.3. If the K_(f) for water = 1.85. What will be the freezing point of solution ?

    A
    `-0.360^(@)C`
    B
    `-0.206^(@)C`
    C
    `+0.480^(@)C`
    D
    `-0.480^(@)C`

    • Similar Questions

    Explore conceptually related problems

    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:-

    In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3 . Taking k_(f) for water as 1.85 the freezing point of the solution will be nearest to-

    The ionization constant of phenol is 1.0xx10^(-10) .What is the concentration of phenolate ion is 0.05 M solution of phenol ? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate ?

    The molal boiling point constant of water is 0.573^(@)C kg "mole"^(-1) . When 0.1 mole of glucose is dissolved in 1000g of water, the solution boils under atmospheric pressure at:

    At certain Hill-station pure water boils at 99.725^(@)C . If K_(b) for water is 0.513^(@)C kg mol^(-1) , the boiling point of 0.69m solution of urea will be:

    Home
    Profile