`PtCl_(4).6H_(2)O`can exist as hydrated complex `1` molal aq.solution has depression in freezing point of `3.72^(@)C`Assume `100%` ionisation and `K_(f)(H_(2)O=1.86^(@)mol^(-1))kg` then complex is
A
`[pt(H_(2)O_(6)]Cl_(4)`
B
`[Pt(H_(2)O)_(5)Cl]Cl_(2) . 2H_(2)O`
C
`[Pt (H_(2)O)_(3)Cl_(3)]Cl. 3H_(2)O`
D
`[Pt (H_(2)O)_(2)Cl_(4)]Cl. 4H_(2)O`
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c
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NARENDRA AWASTHI|Exercise Level 1 (Q.62 To Q.91)|1 Videos
PtCl_(4). 6H_(2)O can exist as hydrated complex 1 molal aq. Solution has depression in freezing point of 3.72^(@) . Assume 100% ionisation and K_(f)(H_(2)O) = 1.86^(@) mol^(-1) kg , then complex is :
PtCl_(4).6H_(2)O can exist as a hydrated complex. 1 m aqueous solution has the depression in freezing point of 3.72^(@) . Assume 100% ionization and K_(f)(H_(2)O)=1.86^(@)mol^(-1) kg , then the complex is
Compound PdCl_(4). 6H_(2)O is a hydrated complex, 1 molal aqueous solution of it has freezing point 269.28 K. Assuming 100% ionization of complex, calculate the molecular formula of the complex ( K_(f) for water = 1.86 K mol^(-1) )
Calculate depression of freezing point for 0.56 molal aq. Solution of KCl. (Given : K_f(H_(2)O) = 1.8 kg mol^(-1) ).
Compound PdCl_(4).6H_(2)O is a hydrated complex, 1 m aqueous solution of it has freezing point 269.28 K . Assuming 100% ionization of complex, calculate the number of ions furnished by complex in the solution.
0.1 mole of sugar is dissolved in 250 g of water. The freezing point of the solution is [K_(f) "for" H_(2)O = 1.86^(@)C "molal"^(-1)]
An aqueous solution of urea has freezing point of -0.52^(@)C . If molarity and molality are same and K'_(f) for H_(2)O = 1.86 K "molality"^(-1) the osmotic pressure of solution would be:
NARENDRA AWASTHI-DILUTE SOLUTION-Level 3 - Match The Column
