An aqueous solution containing ionic salt having molality equal to 0.1892 freezes at -0.704° C. The van't Hoff factor of the ionic salt will be equal to ( K f =1.86 Km -1)

Question

An aqueous solution containing ionic salt having molality equal to 0.1892 freezes at -0.704° C. The van't Hoff factor of the ionic salt will be equal to ( K f =1.86 Km -1) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ T f = i × K f × m 0-(-0.704)= i × 1.86 × 0.1892 i =(0.704/1.86 × 0.1892) ≈ 2

Q. An aqueous solution containing ionic salt having molality equal to $0.1892$ freezes at $- 0.70 4^{\circ} C$ . The van't Hoff factor of the ionic salt will be equal to _____ $(K_{f} = 1.86 K m^{- 1})$

$Δ T_{f} = i \times K_{f} \times m$
$0 - (- 0.704) = i \times 1.86 \times 0.1892$
$i = \frac{0.704}{1.86 \times 0.1892} \approx 2$

Q. An aqueous solution containing ionic salt having molality equal to 0.1892 freezes at −0.704∘C. The van't Hoff factor of the ionic salt will be equal to _____ (Kf​=1.86Km−1)

ΔTf​=i×Kf​×m 0−(−0.704)=i×1.86×0.1892 i=1.86×0.18920.704​≈2

Q. An aqueous solution containing ionic salt having molality equal to $0.1892$ freezes at $- 0.70 4^{\circ} C$ . The van't Hoff factor of the ionic salt will be equal to _____ $(K_{f} = 1.86 K m^{- 1})$

$Δ T_{f} = i \times K_{f} \times m$
$0 - (- 0.704) = i \times 1.86 \times 0.1892$
$i = \frac{0.704}{1.86 \times 0.1892} \approx 2$