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A dilute solution containing 2.5g of a non volatile non electrolytic solute in 100 g water, the elevation in boiling point at 1atm pressure is 2 degree Celsius. assuming concentration of the solute is much lower than the concentration of the solvent, the vapour pressure (mm of Hg) of the solution is (ebuliosscopic constant= 0.76K kg/mol) A) 724 B) 740 C) 736 C) 718?
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A dilute solution containing 2.5g of a non volatile non electrolytic s...
Frst use ∆Tb= i kb m..thn find molality ,m,frm here..thn use m to find moles of solute.. u will get moles= 5/19..thn use Po--Ps/Po= n/n+N.. as its given solute is very less as conpared to solvent,so n is ignored in denominator..we can take Po--Ps/Po=n/N..take value of Po= 760 mmHg (bcs it's 1 atm given)..now solve..u will get Ps= 724..
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A dilute solution containing 2.5g of a non volatile non electrolytic s...
Calculation of Vapour Pressure of the Solution:
- The boiling point elevation can be calculated using the formula:
ΔTb = i * Kf * m
Where ΔTb is the boiling point elevation, i is the van't Hoff factor (1 for non-electrolytes), Kf is the ebullioscopic constant, and m is the molality of the solution.
- Given that the elevation in boiling point is 2 degrees Celsius and Kf is 0.76 K kg/mol, we can calculate the molality of the solution:
2 = 1 * 0.76 * (2.5/18)
= 0.1056 mol/kg
- Now, we need to calculate the vapor pressure of the solution using the formula for boiling point elevation:
ΔTb = Tb(solvent) - Tb(solution)
2 = 100 - Tb(solution)
Tb(solution) = 98 degrees Celsius
- The vapor pressure of the solution can be calculated using the Clausius-Clapeyron equation:
ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)
Where P1 is the vapor pressure of the pure solvent, P2 is the vapor pressure of the solution, ΔHvap is the enthalpy of vaporization, R is the gas constant, T1 is the boiling point of the pure solvent, and T2 is the boiling point of the solution.
- By substituting the values into the equation, we can solve for P2:
ln(P2/760) = 40.79 * (1/373 - 1/371)
P2 = 740 mmHg
Therefore, the vapor pressure of the solution is 740 mmHg. Hence, the answer is B) 740 mmHg.
- The boiling point elevation can be calculated using the formula:
ΔTb = i * Kf * m
Where ΔTb is the boiling point elevation, i is the van't Hoff factor (1 for non-electrolytes), Kf is the ebullioscopic constant, and m is the molality of the solution.
- Given that the elevation in boiling point is 2 degrees Celsius and Kf is 0.76 K kg/mol, we can calculate the molality of the solution:
2 = 1 * 0.76 * (2.5/18)
= 0.1056 mol/kg
- Now, we need to calculate the vapor pressure of the solution using the formula for boiling point elevation:
ΔTb = Tb(solvent) - Tb(solution)
2 = 100 - Tb(solution)
Tb(solution) = 98 degrees Celsius
- The vapor pressure of the solution can be calculated using the Clausius-Clapeyron equation:
ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)
Where P1 is the vapor pressure of the pure solvent, P2 is the vapor pressure of the solution, ΔHvap is the enthalpy of vaporization, R is the gas constant, T1 is the boiling point of the pure solvent, and T2 is the boiling point of the solution.
- By substituting the values into the equation, we can solve for P2:
ln(P2/760) = 40.79 * (1/373 - 1/371)
P2 = 740 mmHg
Therefore, the vapor pressure of the solution is 740 mmHg. Hence, the answer is B) 740 mmHg.
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A dilute solution containing 2.5g of a non volatile non electrolytic solute in 100 g water, the elevation in boiling point at 1atm pressure is 2 degree Celsius. assuming concentration of the solute is much lower than the concentration of the solvent, the vapour pressure (mm of Hg) of the solution is (ebuliosscopic constant= 0.76K kg/mol) A) 724 B) 740 C) 736 C) 718?
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A dilute solution containing 2.5g of a non volatile non electrolytic solute in 100 g water, the elevation in boiling point at 1atm pressure is 2 degree Celsius. assuming concentration of the solute is much lower than the concentration of the solvent, the vapour pressure (mm of Hg) of the solution is (ebuliosscopic constant= 0.76K kg/mol) A) 724 B) 740 C) 736 C) 718? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A dilute solution containing 2.5g of a non volatile non electrolytic solute in 100 g water, the elevation in boiling point at 1atm pressure is 2 degree Celsius. assuming concentration of the solute is much lower than the concentration of the solvent, the vapour pressure (mm of Hg) of the solution is (ebuliosscopic constant= 0.76K kg/mol) A) 724 B) 740 C) 736 C) 718? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A dilute solution containing 2.5g of a non volatile non electrolytic solute in 100 g water, the elevation in boiling point at 1atm pressure is 2 degree Celsius. assuming concentration of the solute is much lower than the concentration of the solvent, the vapour pressure (mm of Hg) of the solution is (ebuliosscopic constant= 0.76K kg/mol) A) 724 B) 740 C) 736 C) 718?.
A dilute solution containing 2.5g of a non volatile non electrolytic solute in 100 g water, the elevation in boiling point at 1atm pressure is 2 degree Celsius. assuming concentration of the solute is much lower than the concentration of the solvent, the vapour pressure (mm of Hg) of the solution is (ebuliosscopic constant= 0.76K kg/mol) A) 724 B) 740 C) 736 C) 718? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A dilute solution containing 2.5g of a non volatile non electrolytic solute in 100 g water, the elevation in boiling point at 1atm pressure is 2 degree Celsius. assuming concentration of the solute is much lower than the concentration of the solvent, the vapour pressure (mm of Hg) of the solution is (ebuliosscopic constant= 0.76K kg/mol) A) 724 B) 740 C) 736 C) 718? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A dilute solution containing 2.5g of a non volatile non electrolytic solute in 100 g water, the elevation in boiling point at 1atm pressure is 2 degree Celsius. assuming concentration of the solute is much lower than the concentration of the solvent, the vapour pressure (mm of Hg) of the solution is (ebuliosscopic constant= 0.76K kg/mol) A) 724 B) 740 C) 736 C) 718?.
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