The boiling point of 0.001 M aqueous solutions of NaCl,Na2SO4,K3PPO4 and CH3COOH should follows the order.

Question

The boiling point of 0.001 M aqueous solutions of NaCl,Na2SO4,K3PPO4 and CH3COOH should follows the order. (A) CH3COOH< NaCl< Na2SO4< K3PO4 (B) NaCl< Na2SO4< K3PO4< CH3 COOH (C) CH3COOH< K3PO4< Na2SO4 < NaCl (D) CH3COOH < K3PO4< NaCl < Na2SO4

Tardigrade Admin · Accepted Answer

Elevation in boiling point, Δ Tb = i × Kb × m.( Kb is constant for a solvent) If molality (m) is same,Δ Tb propto van’t Hoff factor (i)CH3 COOH only partially dissociates, ∴ i < 2 NaCl, Na2SO4 and K3PO4 almost completely dissociate to produce 2, 3 and 4 ions per formula respectively, ∴ i ≈ 2, 3, and 4 respectively ∴ The order of i = the order of boiling points = CH3COOH < NaCl < Na2SO4 < K3PO4

Q. The boiling point of $0.001 M$ aqueous solutions of $N a Cl, N a_{2} S O_{4}, K_{3} PP O_{4}$ and $C H_{3} COO H$ should follows the order.

$C H_{3} COO H < N a Cl < N a_{2} S O_{4} < K_{3} P O_{4}$

$N a Cl < N a_{2} S O_{4} < K_{3} P O_{4} < C H_{3} COO H$

$C H_{3} COO H < K_{3} P O_{4} < N a_{2} S O_{4} < N a Cl$

$C H_{3} COO H < K_{3} P O_{4} < N a Cl < N a_{2} S O_{4}$

Q. The boiling point of 0.001M aqueous solutions of NaCl,Na2​SO4​,K3​PPO4​ and CH3​COOH should follows the order.

CH3​COOH<NaCl<Na2​SO4​<K3​PO4​

NaCl<Na2​SO4​<K3​PO4​<CH3​COOH

CH3​COOH<K3​PO4​<Na2​SO4​<NaCl

CH3​COOH<K3​PO4​<NaCl<Na2​SO4​

Q. The boiling point of $0.001 M$ aqueous solutions of $N a Cl, N a_{2} S O_{4}, K_{3} PP O_{4}$ and $C H_{3} COO H$ should follows the order.

$C H_{3} COO H < N a Cl < N a_{2} S O_{4} < K_{3} P O_{4}$

$N a Cl < N a_{2} S O_{4} < K_{3} P O_{4} < C H_{3} COO H$

$C H_{3} COO H < K_{3} P O_{4} < N a_{2} S O_{4} < N a Cl$

$C H_{3} COO H < K_{3} P O_{4} < N a Cl < N a_{2} S O_{4}$