Y g of a non-volatile solute of molar mass M is dissolved in 250 g of benzene. If Kb is molal elevation constant, the value of Δ T is given by:

Question

Y g of a non-volatile solute of molar mass M is dissolved in 250 g of benzene. If Kb is molal elevation constant, the value of Δ T is given by: (A) 2Y Kb/M (B) 4 Y Kb/M (C) 2Y Kb/4 M (D) 2.5Y Kb/M

Tardigrade Admin · Accepted Answer

Δ T=Kb×(wB×1000/mB× wA) =Kb ×(Y×1000/M×250) =(4Kb Y/M)

Q. Y g of a non-volatile solute of molar mass M is dissolved in 250 g of benzene. If $K_{b}$ is molal elevation constant, the value of $Δ T$ is given by:

$2 Y K_{b} / M$

$4 Y K_{b} / M$

$2 Y K_{b} /4 M$

$2.5 Y K_{b} / M$

$Δ T = K_{b} \times \frac{w _{B} \times 1000}{m _{B} \times w _{A}}$
$= K_{b} \times \frac{Y \times 1000}{M \times 250}$
$= \frac{4 K _{b} Y}{M}$

Q. Y g of a non-volatile solute of molar mass M is dissolved in 250 g of benzene. If Kb​ is molal elevation constant, the value of ΔT is given by:

2YKb​/M

4YKb​/M

2YKb​/4M

2.5YKb​/M