Home
Class 12
CHEMISTRY
0.002m aqueous solution of an ionic comp...

`0.002m` aqueous solution of an ionic compound `Co(NH_(3))_(5)(NO_(2))CI` freezes at `-0.00732^(@)C`.Number of moles of ions which 1 mole of ionic compound produces in water will be `(K_(f) = 1.86^(@)C//m)`

A

3

B

4

C

1

D

2

Text Solution

Verified by Experts

The correct Answer is:
D

The number of moles produced by 1 mole of ionic comppound.
` Delta T _(f) = ik_(f) m, I = Delta T_(f)//K_(f) m`
`Delta T_(f) = 0 - (- 0.00732)`
` = 0.00732^(@)`
`k_(f) = 1.86^(@)C//m`
m = 0.0020
` I = (0.00732)/(1.86 xx 0.0020)`
` = 1.968 ~~ 2`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • SOLUTIONS

    MODERN PUBLICATION|Exercise COMPETITION FILE (C. MULTIPLE CHOICE QUESTIONS(MCQ))|10 Videos
  • SOLUTIONS

    MODERN PUBLICATION|Exercise COMPETITION FILE (D. MULTIPLE CHOICE QUESTIONS(MCQ))|13 Videos
  • SOLUTIONS

    MODERN PUBLICATION|Exercise COMPETITION FILE (A. MULTIPLE CHOICE QUESTIONS(MCQ))|26 Videos
  • SOLID STATE

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|13 Videos
  • SURFACE CHEMISTRY

    MODERN PUBLICATION|Exercise COMPETITION FILE (OBJECTIVE TYPE QUESTIONS) (MATCHING LIST TYPE QUESTIONS)|2 Videos

Similar Questions

Explore conceptually related problems

A 0.0020 m aqueous solution of an ionic compound Co(NH_(3))_(5) (NO_(2))Cl freezes at -0.00744^(@)C . The number of moles of ions which 1 mole of ionic compound produces on being dissolve in water is (K_(f)=-1.86^(@)C//m)

0.002 molal aqueous solution of an ionic compound with molecular formula Co(NH_(3))_(5)(NO_(2))Cl freezes at -0.00744^(@)C . How many moles of ions does 3 moles of the salt produce on being dissolved in water? [Given K_(f) of water=1.86 K kg mol^(-1) ]

Knowledge Check

  • A 0.0020m aqueous solution of an ionic compound Co(NH_(3))_(5)(NO_(2))Cl freezes at -0.00732^(@)C . Number of moles of ions which 1 mol of ionic comound produces on being dissolved in water be (K_(f)=-1.86^(@)C//m)

    A
    2
    B
    3
    C
    4
    D
    1
  • A 0.0020 in aqueous solution of an ionic compound [CO(NH_3)_5(NO_2)]Cl freezes at -0.00732^@C . Number of moles of ions which I mol of ionic compound produces on being dissolved in water will be (K_f = -1.86^@C//m)

    A
    3
    B
    4
    C
    1
    D
    2
  • A 0.0020 m aqueous solution of an ionic compound [CO(NH_3)_5 (NO_2)]CI freezes at -0.00732 ""^@C . Number of moles of ions in which I mol of ionic compound produces on being dissolved in water will be (K_f = -1.86^@C//m)

    A
    3
    B
    4
    C
    1
    D
    2
  • Similar Questions

    Explore conceptually related problems

    A complex is prepared by mixing CoCl_(3) and NH_(3) in the molar ratio of 1 : 4 . 0.1 M solution of this complex was found to freeze at -0.372^(@)C . What is the formula of the complex ? Given that molal depression constant of water (K_(f))=1.86^(@)C//m

    A 0.0020 m aqueous solution os an ionic compound [Co(NH_(3))_(5)(NO_(2))] CI freezes at -0.0732^(@)C . Number of moles of ions which 1 mole of an ionic compound produces on being dissolved in water will be:

    The freezing point of 0.05 m solution of glucose in water is (K1 = 1.86°C m^(-1) )

    The freezing point of a 0.05 molal solution of a non-electrolyte in water is [K_(f)=1.86K//m]

    1 xx 10^(-3) m solution of Pt(NH_(3))_(4)Cl_(4) in H_(2)O shows depression in freezing point of 0.0054^(@)C . The formula of the compound will be [Given K_(f) (H_(2)O) = 1.86^(@)C m^(-1) ]