The boiling point of pure benzene is 80.1° C. When 2 g of a non-volatile solute (molar mass =60 g mol -1 ) was dissolved in 100 g of benzene. The solution will boil at ° C. ( K b . for benzene .=2.53 K m -1)

Question

The boiling point of pure benzene is 80.1° C. When 2 g of a non-volatile solute (molar mass =60 g mol -1 ) was dissolved in 100 g of benzene. The solution will boil at ° C. ( K b . for benzene .=2.53 K m -1) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ T b = K b ( W B / M B W A ) =2.53 × (2/60 × 0.1) =0.84 K or 0.84° C Δ T b = T b - T b ° ∴ T b = T b °+Δ T b =80.1+0.84 =80.94 ° C

Q. The boiling point of pure benzene is $80. 1^{\circ} C$ . When $2 g$ of a non-volatile solute (molar mass $= 60 g m o l^{- 1}$ ) was dissolved in $100 g$ of benzene. The solution will boil at_____ $^{\circ} C$ .
$(K_{b}$ for benzene $= 2.53 K m^{- 1})$

$Δ T_{b} = K_{b} \frac{W _{B}}{M _{B} W _{A}}$
$= 2.53 \times \frac{2}{60 \times 0.1}$
$= 0.84 K$
or $0.8 4^{\circ} C$
$Δ T_{b} = T_{b} - T_{b}^{\circ}$
$∴ T_{b} = T_{b}^{\circ} + Δ T_{b}$
$= 80.1 + 0.84$
$= 80.94^{\circ} C$

Q. The boiling point of pure benzene is 80.1∘C. When 2g of a non-volatile solute (molar mass =60gmol−1 ) was dissolved in 100g of benzene. The solution will boil at_____ ∘C. (Kb​ for benzene =2.53Km−1)

ΔTb​=Kb​MB​WA​WB​​ =2.53×60×0.12​ =0.84K or 0.84∘C ΔTb​=Tb​−Tb∘​ ∴Tb​=Tb∘​+ΔTb​ =80.1+0.84 =80.94∘C

Q. The boiling point of pure benzene is $80. 1^{\circ} C$ . When $2 g$ of a non-volatile solute (molar mass $= 60 g m o l^{- 1}$ ) was dissolved in $100 g$ of benzene. The solution will boil at_____ $^{\circ} C$ .
$(K_{b}$ for benzene $= 2.53 K m^{- 1})$

$Δ T_{b} = K_{b} \frac{W _{B}}{M _{B} W _{A}}$
$= 2.53 \times \frac{2}{60 \times 0.1}$
$= 0.84 K$
or $0.8 4^{\circ} C$
$Δ T_{b} = T_{b} - T_{b}^{\circ}$
$∴ T_{b} = T_{b}^{\circ} + Δ T_{b}$
$= 80.1 + 0.84$
$= 80.94^{\circ} C$