An aqueous dilute solution containing non-volatile solute boils at 100.52° C. What is the molality of solution? (Kb=0.52 kg mol -1 K., boiling temperature of water =100° C )

Question

An aqueous dilute solution containing non-volatile solute boils at 100.52° C. What is the molality of solution? (Kb=0.52 kg mol -1 K., boiling temperature of water =100° C ) (A) 0.1 m (B) 0.01 m (C) 0.001 m (D) 1.0 m

Tardigrade Admin · Accepted Answer

Δ Tb=Kb ⋅ m Tb-Tb°=Kb ⋅ m where, Kb= molal elevation constant m= molality (100.052-100.00) =(0.52)( m ) 0.052 =0.52 × m m =0.1

Q. An aqueous dilute solution containing non-volatile solute boils at $100.5 2^{\circ} C$ . What is the molality of solution? $(K_{b} = 0.52 k g m o l^{- 1} K$ , boiling temperature of water $= 10 0^{\circ} C$ )

0.1 m

0.01 m

0.001 m

1.0 m

$Δ T_{b} = K_{b} \cdot m$

$T_{b} - T_{b}^{\circ} = K_{b} \cdot m$

where, $K_{b} =$ molal elevation constant

$m =$ molality

$(100.052 - 100.00) = (0.52) (m)$

$0.052 = 0.52 \times m$

$m = 0.1$

Q. An aqueous dilute solution containing non-volatile solute boils at 100.52∘C. What is the molality of solution? (Kb​=0.52kgmol−1K, boiling temperature of water =100∘C )