When 1.5 mol of CuCl 2 ⋅ 2 H 2 O is dissolved in enough water to make 1.0 L of solution Given: Kf CuCl ⊕=1.0(Kf. is the formation constant of .CuCl ⊕) [ Cu 2+] in solution is

Question

When 1.5 mol of CuCl 2 ⋅ 2 H 2 O is dissolved in enough water to make 1.0 L of solution Given: Kf CuCl ⊕=1.0(Kf. is the formation constant of .CuCl ⊕) [ Cu 2+] in solution is (A) 1.0 M (B) 0.5 M (C) 2.0 M (D) None

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/0f91d836377d15ca552653db60802321-.png" /> Kf=([ CuCl ⊕]/[ Cu 2+][ Cl ominus])=1.0 ⇒ (x/(1.5-x)(3.0-x))=1.0 ⇒ x=4.5-4.5 x+x2 ⇒ x2-5.5 x+4.5=0 x=(5.5 ± √(5.5)2-4(4.5)/2), x=1.0 or 4.5 [ Cu 2+]=0.5 M

Q. When $1.5 m o l$ of $C u C l_{2} \cdot 2 H_{2} O$ is dissolved in enough water to make $1.0 L$ of solution Given: $K_{f} C u C l^{\oplus} = 1.0 (K_{f}$ is the formation constant of $C u C l^{\oplus})$
$[C u^{2 +}]$ in solution is

$1.0 M$

$0.5 M$

$2.0 M$

None

$K_{f} = \frac{[ C u C l ^{\oplus} ]}{[ C u ^{2 +} ] [ C l ^{⊖} ]} = 1.0$
$\Rightarrow \frac{x}{( 1.5 - x ) ( 3.0 - x )} = 1.0$
$\Rightarrow x = 4.5 - 4.5 x + x^{2}$
$\Rightarrow x^{2} - 5.5 x + 4.5 = 0$
$x = \frac{5.5 \pm ( 5.5 ) ^{2} - 4 ( 4.5 )}{2}, x = 1.0$ or $4.5$
$[C u^{2 +}] = 0.5 M$

Q. When 1.5mol of CuCl2​⋅2H2​O is dissolved in enough water to make 1.0L of solution Given: Kf​CuCl⊕=1.0(Kf​ is the formation constant of CuCl⊕) [Cu2+] in solution is

1.0M

0.5M

2.0M