A $0.25$ molal aqueous solution of $C O {({N H}_{3})}_{4} {C l}_{3}$ freezes at $- {0.93}^{o} C$ . Calculate the number of stereoisomers of above coordination compound. $K_{t}$ of $H_{2} O = 1.86 K / m o l a l$ .

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Solution

$m = 0.25$
$C O (N H_{3})_{4} C l_{3}$
$\Rightarrow △ T_{f} = 0.93$
$\Rightarrow △ T_{f} = i k_{f} m$
$\Rightarrow \frac{0.93}{1.86 \times 0.25} = i$
$\Rightarrow i = 2$
$∴ 2$ stereoisomers are possible.

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A 0.25 molal aqueous solution of CO(NH3)4Cl3 freezes at −0.93oC. Calculate the number of stereoisomers of above coordination compound. Kt of H2O=1.86K/molal.

m=0.25CO(NH3)4Cl3⇒△Tf=0.93⇒△Tf=ikfm⇒0.931.86×0.25=i⇒i=2∴2 stereoisomers are possible.

A $0.25$ molal aqueous solution of $C O {({N H}_{3})}_{4} {C l}_{3}$ freezes at $- {0.93}^{o} C$ . Calculate the number of stereoisomers of above coordination compound. $K_{t}$ of $H_{2} O = 1.86 K / m o l a l$ .

$m = 0.25$
$C O (N H_{3})_{4} C l_{3}$
$\Rightarrow △ T_{f} = 0.93$
$\Rightarrow △ T_{f} = i k_{f} m$
$\Rightarrow \frac{0.93}{1.86 \times 0.25} = i$
$\Rightarrow i = 2$
$∴ 2$ stereoisomers are possible.