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An elevation in boiling point of a solution of 10 gram of solute( molar mass = 100) in 100 gram of water is ∆ Tb the ebullioscopic constant of water is. a) 10 b) 10∆Tb c) ∆Tb d) ∆Tb/10?
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An elevation in boiling point of a solution of 10 gram of solute( mola...
Boiling point elevation and ebullioscopic constant
Boiling Point Elevation:
When a non-volatile solute is added to a solvent, the boiling point of the solution increases. This phenomenon is called boiling point elevation. The elevation in boiling point is directly proportional to the molality of the solution.
Ebullioscopic Constant:
The ebullioscopic constant is a proportionality constant that relates the elevation in boiling point to the molality of the solution. The value of the ebullioscopic constant depends on the solvent used. For water, the value of the ebullioscopic constant is 0.52 °C/m.
Answer to the question:
The elevation in boiling point of a solution of 10 gram of solute(molar mass = 100) in 100 gram of water is denoted by ΔTb. The ebullioscopic constant of water is given by 0.52 °C/m.
Therefore, the correct answer is option (c) ΔTb.
Explanation:
The elevation in boiling point of the solution is directly proportional to the molality of the solution and the ebullioscopic constant of the solvent. The molality of the solution is given by the formula:
molality (m) = (moles of solute) / (mass of solvent in kg)
Here, the mass of solute is given as 10 g and the molar mass of the solute is given as 100 g/mol. Therefore, the number of moles of solute is:
moles of solute = (mass of solute) / (molar mass of solute)
= (10 g) / (100 g/mol)
= 0.1 mol
The mass of solvent is given as 100 g. Therefore, the mass of solvent in kg is:
mass of solvent = (100 g) / (1000 g/kg) = 0.1 kg
Using the above values, the molality of the solution is calculated as:
molality (m) = (0.1 mol) / (0.1 kg) = 1 mol/kg
The elevation in boiling point is given by the formula:
ΔTb = Kbm
Here, K is the ebullioscopic constant of the solvent and bm is the molality of the solution. Substituting the values of K and bm, we get:
ΔTb = (0.52 °C/m) x (1 mol/kg) = 0.52 °C
Therefore, the correct answer is option (c) ΔTb.
Boiling Point Elevation:
When a non-volatile solute is added to a solvent, the boiling point of the solution increases. This phenomenon is called boiling point elevation. The elevation in boiling point is directly proportional to the molality of the solution.
Ebullioscopic Constant:
The ebullioscopic constant is a proportionality constant that relates the elevation in boiling point to the molality of the solution. The value of the ebullioscopic constant depends on the solvent used. For water, the value of the ebullioscopic constant is 0.52 °C/m.
Answer to the question:
The elevation in boiling point of a solution of 10 gram of solute(molar mass = 100) in 100 gram of water is denoted by ΔTb. The ebullioscopic constant of water is given by 0.52 °C/m.
Therefore, the correct answer is option (c) ΔTb.
Explanation:
The elevation in boiling point of the solution is directly proportional to the molality of the solution and the ebullioscopic constant of the solvent. The molality of the solution is given by the formula:
molality (m) = (moles of solute) / (mass of solvent in kg)
Here, the mass of solute is given as 10 g and the molar mass of the solute is given as 100 g/mol. Therefore, the number of moles of solute is:
moles of solute = (mass of solute) / (molar mass of solute)
= (10 g) / (100 g/mol)
= 0.1 mol
The mass of solvent is given as 100 g. Therefore, the mass of solvent in kg is:
mass of solvent = (100 g) / (1000 g/kg) = 0.1 kg
Using the above values, the molality of the solution is calculated as:
molality (m) = (0.1 mol) / (0.1 kg) = 1 mol/kg
The elevation in boiling point is given by the formula:
ΔTb = Kbm
Here, K is the ebullioscopic constant of the solvent and bm is the molality of the solution. Substituting the values of K and bm, we get:
ΔTb = (0.52 °C/m) x (1 mol/kg) = 0.52 °C
Therefore, the correct answer is option (c) ΔTb.
Community Answer
An elevation in boiling point of a solution of 10 gram of solute( mola...
Option C. ∆Tb=kb×m. kb= ebullioscopic constant. m=molality.
m=w/gmwt ×1000/solvent (g)= (10/100)×(1000/100)=1.
kb=∆Tb/m. therefore kb=∆Tb.
m=w/gmwt ×1000/solvent (g)= (10/100)×(1000/100)=1.
kb=∆Tb/m. therefore kb=∆Tb.
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An elevation in boiling point of a solution of 10 gram of solute( molar mass = 100) in 100 gram of water is ∆ Tb the ebullioscopic constant of water is. a) 10 b) 10∆Tb c) ∆Tb d) ∆Tb/10?
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An elevation in boiling point of a solution of 10 gram of solute( molar mass = 100) in 100 gram of water is ∆ Tb the ebullioscopic constant of water is. a) 10 b) 10∆Tb c) ∆Tb d) ∆Tb/10? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An elevation in boiling point of a solution of 10 gram of solute( molar mass = 100) in 100 gram of water is ∆ Tb the ebullioscopic constant of water is. a) 10 b) 10∆Tb c) ∆Tb d) ∆Tb/10? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An elevation in boiling point of a solution of 10 gram of solute( molar mass = 100) in 100 gram of water is ∆ Tb the ebullioscopic constant of water is. a) 10 b) 10∆Tb c) ∆Tb d) ∆Tb/10?.
An elevation in boiling point of a solution of 10 gram of solute( molar mass = 100) in 100 gram of water is ∆ Tb the ebullioscopic constant of water is. a) 10 b) 10∆Tb c) ∆Tb d) ∆Tb/10? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An elevation in boiling point of a solution of 10 gram of solute( molar mass = 100) in 100 gram of water is ∆ Tb the ebullioscopic constant of water is. a) 10 b) 10∆Tb c) ∆Tb d) ∆Tb/10? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An elevation in boiling point of a solution of 10 gram of solute( molar mass = 100) in 100 gram of water is ∆ Tb the ebullioscopic constant of water is. a) 10 b) 10∆Tb c) ∆Tb d) ∆Tb/10?.
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