Molal elevation constant of a liquid is:

Question

Molal elevation constant of a liquid is: (A) the elevation in b.p. which would be produced by dissolving one mole of solute in 100 g of solvent (B) the elevation of b.p. which would be produced by dissolving 1 mole solute in 10 g of solvent (C) elevation in b.p. which would be produced by dissolving 1 mole of solute in 1000 g of solvent (D) none of the above

Tardigrade Admin · Accepted Answer

Δ Tb=(1000K'b × w/m.w); if w/m =1, w=1000 g Δ Tb=K'b

Q. Molal elevation constant of a liquid is:

the elevation in b.p. which would be produced by dissolving one mole of solute in 100 g of solvent

the elevation of b.p. which would be produced by dissolving 1 mole solute in 10 g of solvent

elevation in b.p. which would be produced by dissolving 1 mole of solute in 1000 g of solvent

none of the above

$Δ T_{b} = \frac{1000 K _{b}^{'} \times w}{m . w}$ ; if w/m =1,
$w = 1000 g$
$Δ T_{b} = K_{b}^{'}$

Q. Molal elevation constant of a liquid is:

the elevation in b.p. which would be produced by dissolving one mole of solute in 100 g of solvent

the elevation of b.p. which would be produced by dissolving 1 mole solute in 10 g of solvent

elevation in b.p. which would be produced by dissolving 1 mole of solute in 1000 g of solvent

none of the above

ΔTb​=m.w1000Kb′​×w​; if w/m =1, w=1000g ΔTb​=Kb′​

$Δ T_{b} = \frac{1000 K _{b}^{'} \times w}{m . w}$ ; if w/m =1,
$w = 1000 g$
$Δ T_{b} = K_{b}^{'}$