A solution of urea (mol. mass 56 g mol -1 ) boils at 100.18° C at the atmospheric pressure. If k f and kb for water are 1.86 and 0.512 K kg mol -1 respectively, the above solution will freeze at

Question

A solution of urea (mol. mass 56 g mol -1 ) boils at 100.18° C at the atmospheric pressure. If k f and kb for water are 1.86 and 0.512 K kg mol -1 respectively, the above solution will freeze at (A) -6.54° C (B)  6.54° C (C)  0.654° C (D)  - 0.654° C

Tardigrade Admin · Accepted Answer

∵ Δ Tf=kf × Molality of solution Δ Tb=kb × Molality of solution or(Δ Tf/Δ Tb)=(kf/kb) Given that Δ Tb=T2-T1=10018-100=0.18 kf for water =1.86 K kg mol -1 kb for water =0.512 K kg mol -1 ∴ (Δ Tf/018) =(1.86/0.512) Δ Tf =(1.86 × 018/0.512) =0.6539 ∼ 0.654 Δ Tf =T1-T2 0.654 =0° C -T2 T2 =-0.654° C (T2 arrow. Freezing point of aqueous urea solution)

Q. A solution of urea (mol. mass $56 g m o l^{- 1}$ ) boils at $100.1 8^{\circ} C$ at the atmospheric pressure. If $k_{f}$ and $k_{b}$ for water are $1.86$ and $0.512 K k g m o l^{- 1}$ respectively, the above solution will freeze at

$- 6.5 4^{\circ} C$

$6.5 4^{\circ} C$

$0.65 4^{\circ} C$

$- 0.65 4^{\circ} C$

Q. A solution of urea (mol. mass 56gmol−1 ) boils at 100.18∘C at the atmospheric pressure. If kf​ and kb​ for water are 1.86 and 0.512Kkgmol−1 respectively, the above solution will freeze at

−6.54∘C

6.54∘C

0.654∘C

−0.654∘C

Q. A solution of urea (mol. mass $56 g m o l^{- 1}$ ) boils at $100.1 8^{\circ} C$ at the atmospheric pressure. If $k_{f}$ and $k_{b}$ for water are $1.86$ and $0.512 K k g m o l^{- 1}$ respectively, the above solution will freeze at

$- 6.5 4^{\circ} C$

$6.5 4^{\circ} C$

$0.65 4^{\circ} C$

$- 0.65 4^{\circ} C$