What is the boiling point of a solution containing 0.456 g of camphor (molar mass =152 ) dissolved in 31.4 g of acetone (. b. .p =56.30° C ), if the molecular elevation constant per 100 g of acetone is 17.2° C ?

Question

What is the boiling point of a solution containing 0.456 g of camphor (molar mass =152 ) dissolved in 31.4 g of acetone (. b. .p =56.30° C ), if the molecular elevation constant per 100 g of acetone is 17.2° C ? (A) 56.46° C (B) 36.56° C (C) 56.14° C (D) 72.52° C

Tardigrade Admin · Accepted Answer

We know that Δ Tb=(100 kb × w/W × m) =(100 × 17.2 × 0.456/31.4 × 152) =0.16° C ∴ Boiling point of solution (Ts)=T°+Δ Tb =56.30+0.16 =56.46° C

Q. What is the boiling point of a solution containing $0.456 g$ of camphor (molar mass $= 152$ ) dissolved in $31.4 g$ of acetone $($ b. $p = 56.3 0^{\circ} C)$ , if the molecular elevation constant per $100 g$ of acetone is $17. 2^{\circ} C ?$

$56.4 6^{\circ} C$

$36.5 6^{\circ} C$

$56.1 4^{\circ} C$

$72.5 2^{\circ} C$

We know that
$Δ T_{b} = \frac{100 k _{b} \times w}{W \times m}$
$= \frac{100 \times 17.2 \times 0.456}{31.4 \times 152}$
$= 0.1 6^{\circ} C$
$∴$ Boiling point of solution
$(T_{s}) = T^{\circ} + Δ T_{b}$
$= 56.30 + 0.16$
$= 56.4 6^{\circ} C$

Q. What is the boiling point of a solution containing 0.456g of camphor (molar mass =152 ) dissolved in 31.4g of acetone ( b. p=56.30∘C), if the molecular elevation constant per 100g of acetone is 17.2∘C?

56.46∘C

36.56∘C

56.14∘C

72.52∘C