Relative decrease in vapour pressure of an aqueous solution containing 2 mol [Cu(NH)3)3Cl] Cl in 3 mol H2O is (1/2) When the given solution reacts with excess of AgNO3 solution the number of moles of AgCl produced is

Question

Relative decrease in vapour pressure of an aqueous solution containing 2 mol [Cu(NH)3)3Cl] Cl in 3 mol H2O is (1/2) When the given solution reacts with excess of AgNO3 solution the number of moles of AgCl produced is (A) 1 mol AgCl (B) 0.25 mol AgCl (C) 2 mol AgCl (D) 0.40 mol AgCl

Tardigrade Admin · Accepted Answer

Let degree of ionisation of the complex,

[C u (N H_{3})_{B} Cl] Cl] Cl

be

α

[C u (N H_{3})_{3} Cl] ⇌ [C u (N H_{3}) Cl]^{+} + C l^{-}

i = 1 + a

\frac{Δ P}{P _{1}^{0}} = \frac{n _{1} ( 1 + α )}{n _{1} ( 1 + α ) + n _{2}}

= \frac{2 ( 1 + α )}{2 ( 1 + α ) + 3} = \frac{1}{2}

α = 0.5

Total

C l^{-}

produced

2 α = 2 \times \frac{1}{2} = 1

mole

∴

1 mole of

A g Cl

is produced

Q. Relative decrease in vapour pressure of an aqueous solution containing 2 mol [Cu(NH)3​)3​Cl]Cl in 3 mol H2​O is 21​ When the given solution reacts with excess of AgNO3​ solution the number of moles of AgCl produced is

1 mol AgCl

0.25 mol AgCl

2 mol AgCl

0.40 mol AgCl

Let degree of ionisation of the complex, [Cu(NH3​)B​Cl]Cl]Cl be α [Cu(NH3​)3​Cl]⇌[Cu(NH3​)Cl]++Cl− i=1+a P10​ΔP​=n1​(1+α)+n2​n1​(1+α)​ =2(1+α)+32(1+α)​=21​ α=0.5 Total Cl− produced 2α=2×21​=1 mole ∴ 1 mole of AgCl is produced

Q. Relative decrease in vapour pressure of an aqueous solution containing $2$ mol $[C u (N H)_{3})_{3} Cl] Cl$ in $3$ mol $H_{2} O$ is $\frac{1}{2}$ When the given solution reacts with excess of $A g N O_{3}$ solution the number of moles of $A g Cl$ produced is