A solution containing 28 g of phosphorus in 315 g CS2 (b.p. 46.3°C) boils at 47.98°C If Kb for CS2 is 2.34 K Kg mol-1 The formula of phosphorus is (at. mass of P = 31)

Question

A solution containing 28 g of phosphorus in 315 g CS2 (b.p. 46.3°C) boils at 47.98°C If Kb for CS2 is 2.34 K Kg mol-1 The formula of phosphorus is (at. mass of P = 31) (A) P6 (B) P4 (C) P3 (D) P2

Tardigrade Admin · Accepted Answer

Δ Tb=m Kb=(w/M) ×(1000/W)× Kb Δ Tb=47.98-46.3=1.68 1.68=(28/M)×(1000/315)×2.38 M=(28×1000×2.38/315×1.68)=125.92 Atomicity = ( textMol.wt./ textAt.wt.)=(125.92/31)=4.02 So. Molecule is =P4

Q. A solution containing 28 g of phosphorus in $315 g C S_{2}$ (b.p. $46. 3^{\circ} C)$ boils at $47.9 8^{\circ} C$ If $K_{b}$ for $C S_{2}$ is $2.34 K K g m o l^{- 1}$ The formula of phosphorus is (at. mass of P = 31)

$P_{6}$

$P_{4}$

$P_{3}$

$P_{2}$

Q. A solution containing 28 g of phosphorus in 315gCS2​ (b.p. 46.3∘C) boils at 47.98∘C If Kb​ for CS2​ is 2.34KKgmol−1 The formula of phosphorus is (at. mass of P = 31)

P6​

P4​

P3​

P2​

ΔTb​=mKb​=Mw​×W1000​×Kb​ ΔTb​=47.98−46.3=1.68 1.68=M28​×3151000​×2.38 M=315×1.6828×1000×2.38​=125.92 Atomicity =At.wt.Mol.wt.​=31125.92​=4.02 So. Molecule is =P4​

Q. A solution containing 28 g of phosphorus in $315 g C S_{2}$ (b.p. $46. 3^{\circ} C)$ boils at $47.9 8^{\circ} C$ If $K_{b}$ for $C S_{2}$ is $2.34 K K g m o l^{- 1}$ The formula of phosphorus is (at. mass of P = 31)

$P_{6}$

$P_{4}$

$P_{3}$

$P_{2}$