A 0.0020 M 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 mole of ionic compound produces on being dissolved in water will be: ( k f =1.86° C / m )-

Question

A 0.0020 M 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 mole of ionic compound produces on being dissolved in water will be: ( k f =1.86° C / m )- (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Δ T f = ik f . m i =(Δ T f / k f ⋅ m ) =(0.00732/1.86 × 0.002)=(0.00732/0.00372) i =2 Compound will be .[ Co ( NH 3)5] NO 5 NO 2] Cl Total possible ions =2

Q. A $0.0020 M$ aqueous solution of an ionic compound $C o (N H_{3})_{5} (N O_{2}) Cl$ freezes at $- 0.0073 2^{\circ} C$ . Number of moles of ions which 1 mole of ionic compound produces on being dissolved in water will be : $(k_{f} = 1.8 6^{\circ} C / m) -$

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$Δ T_{f} = i k_{f} . m$

$i = \frac{Δ T _{f}}{k _{f} \cdot m}$

$= \frac{0.00732}{1.86 \times 0.002} = \frac{0.00732}{0.00372}$

$i = 2$

Compound will be $[C o (N H_{3})_{5}] N O_{5} N O_{2}] Cl$

Total possible ions $= 2$

Q. A 0.0020M aqueous solution of an ionic compound Co(NH3​)5​(NO2​)Cl freezes at −0.00732∘C. Number of moles of ions which 1 mole of ionic compound produces on being dissolved in water will be : (kf​=1.86∘C/m)−

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2

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ΔTf​=ikf​.m i=kf​⋅mΔTf​​ =1.86×0.0020.00732​=0.003720.00732​ i=2 Compound will be [Co(NH3​)5​]NO5​NO2​]Cl Total possible ions =2

Q. A $0.0020 M$ aqueous solution of an ionic compound $C o (N H_{3})_{5} (N O_{2}) Cl$ freezes at $- 0.0073 2^{\circ} C$ . Number of moles of ions which 1 mole of ionic compound produces on being dissolved in water will be : $(k_{f} = 1.8 6^{\circ} C / m) -$

$Δ T_{f} = i k_{f} . m$

$i = \frac{Δ T _{f}}{k _{f} \cdot m}$

$= \frac{0.00732}{1.86 \times 0.002} = \frac{0.00732}{0.00372}$

$i = 2$

Compound will be $[C o (N H_{3})_{5}] N O_{5} N O_{2}] Cl$

Total possible ions $= 2$