The freezing point of a solution containing 2.40 g of a compound in 60.0 g of benzene is 0.10° C lower than that of pure benzene. What is the molecular weight (in gm .mol -1) of the compound? ( K f . is 5.12° C / m for benzene)

Question

The freezing point of a solution containing 2.40 g of a compound in 60.0 g of benzene is 0.10° C lower than that of pure benzene. What is the molecular weight (in gm .mol -1) of the compound? ( K f . is 5.12° C / m for benzene) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Solute - B, Solvent -A Δ T f =( K f × w B × 1000/ m B × w A ) ⇒ 0.10=(5.12 × 2.40 × 1000/ m B × 60) m B =2048 gm mol -1

Q. The freezing point of a solution containing $2.40 g$ of a compound in $60.0 g$ of benzene is $0.1 0^{\circ} C$ lower than that of pure benzene. What is the molecular weight (in gm $m o l^{- 1})$ of the compound? $(K_{f}$ is $5.1 2^{\circ} C / m$ for benzene)

Solute - $B$ , Solvent - $A$
$Δ T_{f} = \frac{K _{f} \times w _{B} \times 1000}{m _{B} \times w _{A}}$
$\Rightarrow 0.10 = \frac{5.12 \times 2.40 \times 1000}{m _{B} \times 60}$
$m_{B} = 2048 g m m o l^{- 1}$

Q. The freezing point of a solution containing 2.40g of a compound in 60.0g of benzene is 0.10∘C lower than that of pure benzene. What is the molecular weight (in gm mol−1) of the compound? (Kf​ is 5.12∘C/m for benzene)

Solute - B, Solvent -A ΔTf​=mB​×wA​Kf​×wB​×1000​ ⇒0.10=mB​×605.12×2.40×1000​ mB​=2048gmmol−1

Q. The freezing point of a solution containing $2.40 g$ of a compound in $60.0 g$ of benzene is $0.1 0^{\circ} C$ lower than that of pure benzene. What is the molecular weight (in gm $m o l^{- 1})$ of the compound? $(K_{f}$ is $5.1 2^{\circ} C / m$ for benzene)

Solute - $B$ , Solvent - $A$
$Δ T_{f} = \frac{K _{f} \times w _{B} \times 1000}{m _{B} \times w _{A}}$
$\Rightarrow 0.10 = \frac{5.12 \times 2.40 \times 1000}{m _{B} \times 60}$
$m_{B} = 2048 g m m o l^{- 1}$