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A 0.0020 m aqueous solution of an ionic compound $Co(NH_3)_5(NO_2)Cl$ freezes at $-0.00732^\circ C$. Number of moles of ions which 1 mole of ionic compound produces on being dissolved in water will be ($K_f = 1.86^\circ C/m$)
A
1
B
2
C
3
D
4
Detailed Solution
$\Delta T_f = i\times K_f\times m$
$\Delta T_f = 0 - (-0.00732) = 0.00732^\circ C$
$i = \frac{\Delta T_f}{K_f\times m} = \frac{0.00732}{1.86\times0.0020} = \frac{0.00732}{0.00372}$
$i = 1.97 \approx 2$
So one mole of the compound gives 2 moles of ions; it ionises as $[Co(NH_3)_5(NO_2)]Cl \rightarrow [Co(NH_3)_5(NO_2)]^+ + Cl^-$
Number of moles of ions = 2
$\Delta T_f = 0 - (-0.00732) = 0.00732^\circ C$
$i = \frac{\Delta T_f}{K_f\times m} = \frac{0.00732}{1.86\times0.0020} = \frac{0.00732}{0.00372}$
$i = 1.97 \approx 2$
So one mole of the compound gives 2 moles of ions; it ionises as $[Co(NH_3)_5(NO_2)]Cl \rightarrow [Co(NH_3)_5(NO_2)]^+ + Cl^-$
Number of moles of ions = 2