If 18 g of glucose is dissolved is 250 g of water then freezing point of the solution will be (Kf of water =1.86 K kg mol -1 )

Question

If 18 g of glucose is dissolved is 250 g of water then freezing point of the solution will be (Kf of water =1.86 K kg mol -1 ) (A) -1.74° C (B) -0.25° C (C) -0.74° C (D) -1.3° C

Tardigrade Admin · Accepted Answer

Δ T f = mK f =(18 / 180/250) × 1000 × 1.86 Δ T f =0.74 T f =-0.74° C

Q. If $18 g$ of glucose is dissolved is $250 g$ of water then freezing point of the solution will be ( $K_{f}$ of water $= 1.86 K k g m o l^{- 1}$ )

$- 1.7 4^{\circ} C$

$- 0.2 5^{\circ} C$

$- 0.7 4^{\circ} C$

$- 1. 3^{\circ} C$

$Δ T_{f} = m K_{f}$
$= \frac{18/180}{250} \times 1000 \times 1.86$
$Δ T_{f} = 0.74$
$T_{f} = - 0.7 4^{\circ} C$

Q. If 18g of glucose is dissolved is 250g of water then freezing point of the solution will be (Kf​ of water =1.86Kkgmol−1 )

−1.74∘C

−0.25∘C

−0.74∘C

−1.3∘C