A 1 M solution of glucose reaches dissociation equilibrium given below. C 6 H 12 O 6 leftharpoons 6 HCHO If the equilibrium constant is 0.167 × 10-22, the concentration of HCHO in the equilibrium is

Question

A 1 M solution of glucose reaches dissociation equilibrium given below. C 6 H 12 O 6 leftharpoons 6 HCHO If the equilibrium constant is 0.167 × 10-22, the concentration of HCHO in the equilibrium is (A) 1.60 × 10-8 M (B) 3.20 × 10-6 M (C) 3.20 × 10-4 M (D) 1.60 × 10-4 M

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/08f32966567b4c2c4c543939649ceb60-.png" /> K =([ HCHO ]6/[ C 6 H 12 O 6])=((6 x )6/(1- x )) ≃ ((6 x )6/1) 0.167 × 10-22=(6 x )6 6 x =(0.167 × 10-22)1 / 6=[0.167 × 102 × 10-24]1 / 6 =[16.7]1 / 6 × 10-4 =1.6 × 10-4 [. because .[1.6]6=[2.56]3=6.4 × 2.56=16.8]

Q. A $1 M$ solution of glucose reaches dissociation equilibrium given below.
$C_{6} H_{12} O_{6} ⇌ 6 H C H O$
If the equilibrium constant is $0.167 \times 1 0^{- 22}$ , the concentration of $H C H O$ in the equilibrium is

$1.60 \times 1 0^{- 8} M$

$3.20 \times 1 0^{- 6} M$

$3.20 \times 1 0^{- 4} M$

$1.60 \times 1 0^{- 4} M$

Q. A 1M solution of glucose reaches dissociation equilibrium given below. C6​H12​O6​⇌6HCHO If the equilibrium constant is 0.167×10−22, the concentration of HCHO in the equilibrium is

1.60×10−8M

3.20×10−6M

3.20×10−4M

1.60×10−4M

K=[C6​H12​O6​][HCHO]6​=(1−x)(6x)6​≃1(6x)6​ 0.167×10−22=(6x)6 6x=(0.167×10−22)1/6=[0.167×102×10−24]1/6 =[16.7]1/6×10−4 =1.6×10−4 [ because [1.6]6=[2.56]3=6.4×2.56=16.8]

Q. A $1 M$ solution of glucose reaches dissociation equilibrium given below.
$C_{6} H_{12} O_{6} ⇌ 6 H C H O$
If the equilibrium constant is $0.167 \times 1 0^{- 22}$ , the concentration of $H C H O$ in the equilibrium is

$1.60 \times 1 0^{- 8} M$

$3.20 \times 1 0^{- 6} M$

$3.20 \times 1 0^{- 4} M$

$1.60 \times 1 0^{- 4} M$