An aqueous solution freezes at -0.186° C (kf=1.86° ; kb=0.512°) What is the elevation in boiling point?

Question

An aqueous solution freezes at -0.186° C (kf=1.86° ; kb=0.512°) What is the elevation in boiling point? (A)  0.186  (B)  0.512  (C)  (0.512/1.86)  (D)  0.0512

Tardigrade Admin · Accepted Answer

Δ Tf=kf'× m 0.186=1.86 × m m=(0.186/1.86)=0.1 mol ∴ Δ Tb=kb × molality Δ Tb=0.512 × 0.1 Δ Tb=0.0512

Q. An aqueous solution freezes at $- 0.18 6^{\circ} C (k_{f} = 1.8 6^{\circ}; k_{b} = 0.51 2^{\circ})$ What is the elevation in boiling point?

$0.186$

$0.512$

$\frac{0.512}{1.86}$

$0.0512$

$Δ T_{f} = k_{f}^{'} \times m$
$0.186 = 1.86 \times m$
$m = \frac{0.186}{1.86} = 0.1 m o l$
$∴ Δ T_{b} = k_{b} \times$ molality
$Δ T_{b} = 0.512 \times 0.1$
$Δ T_{b} = 0.0512$

Q. An aqueous solution freezes at −0.186∘C(kf​=1.86∘;kb​=0.512∘) What is the elevation in boiling point?

0.186

0.512

1.860.512​

0.0512