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A ‘100 proof solution of ethanol in water consists of 50.0 mL of C2H5OH(/)and 50.0 mL of H20 (/), mixed a t1 6 °C . Given,
Density of H20 = 1 g mL-1
Density of C2H,OH = 0.7939 gmL-1
Density mixture = 0.9344 g mL-1
Volume of the solution is
- a)100 mL
- b)105 mL
- c)96 mL
- d)98 mL
Correct answer is option 'C'. Can you explain this answer?
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. (c) Volume can be additive or not depend on the nature of the solutio
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Solution:
Given,
Volume of C2H5OH = 50.0 mL
Volume of H2O = 50.0 mL
Density of H2O = 1 g mL-1
Density of C2H5OH = 0.7939 g mL-1
Density of mixture = 0.9344 g mL-1
Temperature = 16°C
To find: Volume of the solution
We know that,
Density = Mass/Volume
Therefore, Mass = Density × Volume
Let the mass of C2H5OH in the solution be 'm'
Mass of C2H5OH = 0.7939 g mL-1 × 50.0 mL = 39.695 g
Mass of H2O = 1 g mL-1 × 50.0 mL = 50.0 g
Total mass of solution = Mass of C2H5OH + Mass of H2O = 39.695 g + 50.0 g = 89.695 g
Density of mixture = Total mass of solution/Volume of solution
Volume of solution = Total mass of solution/Density of mixture
Volume of solution = 89.695 g/0.9344 g mL-1 = 96.05 mL
Therefore, the volume of the solution is 96 mL (approx).
Hence, the correct answer is option (C) 96 mL.
Given,
Volume of C2H5OH = 50.0 mL
Volume of H2O = 50.0 mL
Density of H2O = 1 g mL-1
Density of C2H5OH = 0.7939 g mL-1
Density of mixture = 0.9344 g mL-1
Temperature = 16°C
To find: Volume of the solution
We know that,
Density = Mass/Volume
Therefore, Mass = Density × Volume
Let the mass of C2H5OH in the solution be 'm'
Mass of C2H5OH = 0.7939 g mL-1 × 50.0 mL = 39.695 g
Mass of H2O = 1 g mL-1 × 50.0 mL = 50.0 g
Total mass of solution = Mass of C2H5OH + Mass of H2O = 39.695 g + 50.0 g = 89.695 g
Density of mixture = Total mass of solution/Volume of solution
Volume of solution = Total mass of solution/Density of mixture
Volume of solution = 89.695 g/0.9344 g mL-1 = 96.05 mL
Therefore, the volume of the solution is 96 mL (approx).
Hence, the correct answer is option (C) 96 mL.
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A ‘100 proof solution of ethanol in water consists of 50.0 mL of C2H5OH(/)and 50.0 mL of H20 (/), mixed a t1 6 °C . Given,Density of H20 = 1 g mL-1Density of C2H,OH = 0.7939 gmL-1Density mixture = 0.9344 g mL-1Volume of the solution isa)100 mLb)105 mLc)96 mLd)98 mLCorrect answer is option 'C'. Can you explain this answer?
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A ‘100 proof solution of ethanol in water consists of 50.0 mL of C2H5OH(/)and 50.0 mL of H20 (/), mixed a t1 6 °C . Given,Density of H20 = 1 g mL-1Density of C2H,OH = 0.7939 gmL-1Density mixture = 0.9344 g mL-1Volume of the solution isa)100 mLb)105 mLc)96 mLd)98 mLCorrect answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A ‘100 proof solution of ethanol in water consists of 50.0 mL of C2H5OH(/)and 50.0 mL of H20 (/), mixed a t1 6 °C . Given,Density of H20 = 1 g mL-1Density of C2H,OH = 0.7939 gmL-1Density mixture = 0.9344 g mL-1Volume of the solution isa)100 mLb)105 mLc)96 mLd)98 mLCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ‘100 proof solution of ethanol in water consists of 50.0 mL of C2H5OH(/)and 50.0 mL of H20 (/), mixed a t1 6 °C . Given,Density of H20 = 1 g mL-1Density of C2H,OH = 0.7939 gmL-1Density mixture = 0.9344 g mL-1Volume of the solution isa)100 mLb)105 mLc)96 mLd)98 mLCorrect answer is option 'C'. Can you explain this answer?.
A ‘100 proof solution of ethanol in water consists of 50.0 mL of C2H5OH(/)and 50.0 mL of H20 (/), mixed a t1 6 °C . Given,Density of H20 = 1 g mL-1Density of C2H,OH = 0.7939 gmL-1Density mixture = 0.9344 g mL-1Volume of the solution isa)100 mLb)105 mLc)96 mLd)98 mLCorrect answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A ‘100 proof solution of ethanol in water consists of 50.0 mL of C2H5OH(/)and 50.0 mL of H20 (/), mixed a t1 6 °C . Given,Density of H20 = 1 g mL-1Density of C2H,OH = 0.7939 gmL-1Density mixture = 0.9344 g mL-1Volume of the solution isa)100 mLb)105 mLc)96 mLd)98 mLCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ‘100 proof solution of ethanol in water consists of 50.0 mL of C2H5OH(/)and 50.0 mL of H20 (/), mixed a t1 6 °C . Given,Density of H20 = 1 g mL-1Density of C2H,OH = 0.7939 gmL-1Density mixture = 0.9344 g mL-1Volume of the solution isa)100 mLb)105 mLc)96 mLd)98 mLCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A ‘100 proof solution of ethanol in water consists of 50.0 mL of C2H5OH(/)and 50.0 mL of H20 (/), mixed a t1 6 °C . Given,Density of H20 = 1 g mL-1Density of C2H,OH = 0.7939 gmL-1Density mixture = 0.9344 g mL-1Volume of the solution isa)100 mLb)105 mLc)96 mLd)98 mLCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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ample number of questions to practice A ‘100 proof solution of ethanol in water consists of 50.0 mL of C2H5OH(/)and 50.0 mL of H20 (/), mixed a t1 6 °C . Given,Density of H20 = 1 g mL-1Density of C2H,OH = 0.7939 gmL-1Density mixture = 0.9344 g mL-1Volume of the solution isa)100 mLb)105 mLc)96 mLd)98 mLCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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