The freezing point of one molal NaCl solution assuming NaCl to be 100% dissociated in water(molal depression constant = 1.86) :

Question

The freezing point of one molal NaCl solution assuming NaCl to be 100% dissociated in water(molal depression constant = 1.86) : (A) -2.72°C (B) -3.72°C (C) 2.72°C (D) 3.72°C

Tardigrade Admin · Accepted Answer

For NaCl i=2 Δ Tf=2 Kf m =2 × 1.86 × 1=3.72 Ts=T-Δ Tf =0-3.72=-3.72 ° C

Q. The freezing point of one molal NaCl solution assuming NaCl to be 100% dissociated in water(molal depression constant = 1.86) :

-2.72 $^{\circ}$ C

-3.72 $^{\circ}$ C

2.72 $^{\circ}$ C

3.72 $^{\circ}$ C

For $N a Cl i = 2$
$Δ T_{f} = 2 K_{f} m$
$= 2 \times 1.86 \times 1 = 3.72$
$T_{s} = T - Δ T_{f}$
$= 0 - 3.72 = - 3.72^{\circ} C$

Q. The freezing point of one molal NaCl solution assuming NaCl to be 100% dissociated in water(molal depression constant = 1.86) :

-2.72∘C

-3.72∘C

2.72∘C

3.72∘C

For NaCli=2 ΔTf​=2Kf​m =2×1.86×1=3.72 Ts​=T−ΔTf​ =0−3.72=−3.72∘C

-2.72 $^{\circ}$ C

-3.72 $^{\circ}$ C

2.72 $^{\circ}$ C

3.72 $^{\circ}$ C

For $N a Cl i = 2$
$Δ T_{f} = 2 K_{f} m$
$= 2 \times 1.86 \times 1 = 3.72$
$T_{s} = T - Δ T_{f}$
$= 0 - 3.72 = - 3.72^{\circ} C$