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)`

Text Solution

Verified by Experts

`DeltaT_(f)=iK_(f)m`
`:.i=(DeltaT_(f))/(K_(f)xxm)=(0.00732)/(1.86xx0.02)=2`
Promotional Banner

Topper's Solved these Questions

  • LIQUID SOLUTION

    FIITJEE|Exercise ASSIGNMENT PROBLEMS (SUBJECTIVE) level-I (SHORT ANSWER TYPE)|7 Videos

  • LIQUID SOLUTION

    FIITJEE|Exercise ASSIGNMENT PROBLEMS (SUBJECTIVE) level-I (FILL IN THE BLANKS)|9 Videos

  • LIQUID SOLUTION

    FIITJEE|Exercise MATCHING TYPE SINGLE CORRECT QUESTION|6 Videos

  • IONIC EQUILIBRIUM

    FIITJEE|Exercise SINGLE INTEGER ANSWER QUESTIONS|4 Videos

  • NUCLEIC ACID AND VITAMIN

    FIITJEE|Exercise ASSIGNMENT PROBLEMS (OBJECTIVE)|15 Videos

Similar Questions

Explore conceptually related problems

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) ]

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) ]