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At 90 degree Celsius if I unit product of water is 1.0 into 10^-12 then d association constant of water is [dH2O = 1g/ml] (1) 1.0 × 10^-14 (2) (1.0 × 10^-12) × 55.5 (3) 1.0 × 10^-12 / 55.5 (4) 1.0 × 10^-14/ 55.5?
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At 90 degree Celsius if I unit product of water is 1.0 into 10^-12 the...
Calculation of Dissociation Constant of Water at 90°C
Formula to calculate dissociation constant of water:
Kw = [H+][OH-]
Given:
Temperature = 90°C = 363 K
Unit product of water = 1.0 x 10^-12
Density of water (dH2O) = 1 g/mL
Molar mass of water (MW) = 18.015 g/mol
Step 1: Calculation of Concentration of Water
The density of water is given as 1 g/mL. Therefore, the concentration of water can be calculated as follows:
1 g of water = 1 mL of water = 1000 mmol of water
Moles of water in 1 mL = 1000/18.015 = 55.49 mmol/mL
Moles of water in 1 L = 55.49 mmol/mL x 1000 mL/L = 55.49 mol/L
Step 2: Calculation of Concentration of H+ and OH-
At 90°C, the ion product of water (Kw) is given as 1.0 x 10^-12. Therefore,
Kw = [H+][OH-] = 1.0 x 10^-12
Since water is neutral, [H+] = [OH-]. Thus,
[H+]^2 = 1.0 x 10^-12
[H+] = [OH-] = sqrt(1.0 x 10^-12) = 1.0 x 10^-6 M
Step 3: Calculation of Dissociation Constant of Water
Using the formula of Kw, we can calculate the dissociation constant of water as follows:
Kw = [H+][OH-] = (1.0 x 10^-6)^2 = 1.0 x 10^-12 M^2
Since Kw = Kc x [H2O], where Kc is the dissociation constant of water and [H2O] is the concentration of water, we can rearrange the equation and solve for Kc as follows:
Kc = Kw/[H2O]
Kc = (1.0 x 10^-12)/(55.49 mol/L)
Kc = 1.8 x 10^-14
Therefore, the dissociation constant of water at 90°C is 1.8 x 10^-14.
Answer: (4) 1.0 x 10^-14/55.5
Formula to calculate dissociation constant of water:
Kw = [H+][OH-]
Given:
Temperature = 90°C = 363 K
Unit product of water = 1.0 x 10^-12
Density of water (dH2O) = 1 g/mL
Molar mass of water (MW) = 18.015 g/mol
Step 1: Calculation of Concentration of Water
The density of water is given as 1 g/mL. Therefore, the concentration of water can be calculated as follows:
1 g of water = 1 mL of water = 1000 mmol of water
Moles of water in 1 mL = 1000/18.015 = 55.49 mmol/mL
Moles of water in 1 L = 55.49 mmol/mL x 1000 mL/L = 55.49 mol/L
Step 2: Calculation of Concentration of H+ and OH-
At 90°C, the ion product of water (Kw) is given as 1.0 x 10^-12. Therefore,
Kw = [H+][OH-] = 1.0 x 10^-12
Since water is neutral, [H+] = [OH-]. Thus,
[H+]^2 = 1.0 x 10^-12
[H+] = [OH-] = sqrt(1.0 x 10^-12) = 1.0 x 10^-6 M
Step 3: Calculation of Dissociation Constant of Water
Using the formula of Kw, we can calculate the dissociation constant of water as follows:
Kw = [H+][OH-] = (1.0 x 10^-6)^2 = 1.0 x 10^-12 M^2
Since Kw = Kc x [H2O], where Kc is the dissociation constant of water and [H2O] is the concentration of water, we can rearrange the equation and solve for Kc as follows:
Kc = Kw/[H2O]
Kc = (1.0 x 10^-12)/(55.49 mol/L)
Kc = 1.8 x 10^-14
Therefore, the dissociation constant of water at 90°C is 1.8 x 10^-14.
Answer: (4) 1.0 x 10^-14/55.5
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At 90 degree Celsius if I unit product of water is 1.0 into 10^-12 then d association constant of water is [dH2O = 1g/ml] (1) 1.0 × 10^-14 (2) (1.0 × 10^-12) × 55.5 (3) 1.0 × 10^-12 / 55.5 (4) 1.0 × 10^-14/ 55.5?
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At 90 degree Celsius if I unit product of water is 1.0 into 10^-12 then d association constant of water is [dH2O = 1g/ml] (1) 1.0 × 10^-14 (2) (1.0 × 10^-12) × 55.5 (3) 1.0 × 10^-12 / 55.5 (4) 1.0 × 10^-14/ 55.5? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about At 90 degree Celsius if I unit product of water is 1.0 into 10^-12 then d association constant of water is [dH2O = 1g/ml] (1) 1.0 × 10^-14 (2) (1.0 × 10^-12) × 55.5 (3) 1.0 × 10^-12 / 55.5 (4) 1.0 × 10^-14/ 55.5? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 90 degree Celsius if I unit product of water is 1.0 into 10^-12 then d association constant of water is [dH2O = 1g/ml] (1) 1.0 × 10^-14 (2) (1.0 × 10^-12) × 55.5 (3) 1.0 × 10^-12 / 55.5 (4) 1.0 × 10^-14/ 55.5?.
At 90 degree Celsius if I unit product of water is 1.0 into 10^-12 then d association constant of water is [dH2O = 1g/ml] (1) 1.0 × 10^-14 (2) (1.0 × 10^-12) × 55.5 (3) 1.0 × 10^-12 / 55.5 (4) 1.0 × 10^-14/ 55.5? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about At 90 degree Celsius if I unit product of water is 1.0 into 10^-12 then d association constant of water is [dH2O = 1g/ml] (1) 1.0 × 10^-14 (2) (1.0 × 10^-12) × 55.5 (3) 1.0 × 10^-12 / 55.5 (4) 1.0 × 10^-14/ 55.5? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 90 degree Celsius if I unit product of water is 1.0 into 10^-12 then d association constant of water is [dH2O = 1g/ml] (1) 1.0 × 10^-14 (2) (1.0 × 10^-12) × 55.5 (3) 1.0 × 10^-12 / 55.5 (4) 1.0 × 10^-14/ 55.5?.
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