319335
Vapour pressure of solution and of pure solvent are \(\mathrm{P}_{1}\) and \(\mathrm{P}_{1}^{0}\) respectively. If \(\dfrac{\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}\) is 0.15 , find the mole dfraction of solute.
1 0.66
2 0.85
3 0.15
4 0.33
Explanation:
\(\dfrac{\mathrm{P}_{1}^{0}-\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}=\mathrm{x}_{2}\) (mole dfraction of solute) \(\therefore \dfrac{\mathrm{P}_{1}^{0}}{\mathrm{P}_{1}^{0}}-\dfrac{\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}=\mathrm{x}_{2}\) \(\therefore \mathrm{x}_{2}=1-0.15=0.85\)
MHTCET - 2021
CHXII02:SOLUTIONS
319336
What is vapour pressure of a solution containing \(0.1 \mathrm{~mol}\) of non-volatile solute dissolved in \(16.2 \mathrm{~g}\) of water? \(\left(\mathrm{p}_{1}^{0}=32 \mathrm{~mm} \mathrm{Hg}\right)\)
319337
Assertion : The vapour pressure of a liquid decreases if some non-volatile solute is dissolved in it. Reason : The relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
The introduction of a non-volatile solute reduces the vapor pressure of a liquid solution by displacing solvent molecules at the surface, hindering their escape into the vapor phase. \({\text{RLVP}},{\text{ }}\frac{{\Delta {{\text{P}}_1}}}{{{\text{P}}_1^0}} = {\chi _2}\,({\text{Raoult's}}\,{\text{law}})\) Relative lowering of vapour pressure is equal to the mole dfraction of solute. So option (2) is correct.
CHXII02:SOLUTIONS
319338
The vapour pressure of a solvent decreases by \(2.5 \mathrm{~mm} \mathrm{Hg}\) by adding a solute. What is the mole dfraction of solute? (Vapour pressure of pure solvent is \(250 \mathrm{~mm} \mathrm{Hg}\) )
1 0.88
2 0.01
3 0.1
4 0.99
Explanation:
\(\Delta {\text{P}} = 2.5\,{\text{mm Hg}},{\text{P}}_1^0 = 250\,{\text{mm Hg}}\) According to Raoult's law, \({{\text{x}}_2} = \frac{{\Delta {\text{P}}}}{{{\text{P}}_1^0}} = \frac{{2.5}}{{250}} = 0.01\)
319335
Vapour pressure of solution and of pure solvent are \(\mathrm{P}_{1}\) and \(\mathrm{P}_{1}^{0}\) respectively. If \(\dfrac{\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}\) is 0.15 , find the mole dfraction of solute.
1 0.66
2 0.85
3 0.15
4 0.33
Explanation:
\(\dfrac{\mathrm{P}_{1}^{0}-\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}=\mathrm{x}_{2}\) (mole dfraction of solute) \(\therefore \dfrac{\mathrm{P}_{1}^{0}}{\mathrm{P}_{1}^{0}}-\dfrac{\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}=\mathrm{x}_{2}\) \(\therefore \mathrm{x}_{2}=1-0.15=0.85\)
MHTCET - 2021
CHXII02:SOLUTIONS
319336
What is vapour pressure of a solution containing \(0.1 \mathrm{~mol}\) of non-volatile solute dissolved in \(16.2 \mathrm{~g}\) of water? \(\left(\mathrm{p}_{1}^{0}=32 \mathrm{~mm} \mathrm{Hg}\right)\)
319337
Assertion : The vapour pressure of a liquid decreases if some non-volatile solute is dissolved in it. Reason : The relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
The introduction of a non-volatile solute reduces the vapor pressure of a liquid solution by displacing solvent molecules at the surface, hindering their escape into the vapor phase. \({\text{RLVP}},{\text{ }}\frac{{\Delta {{\text{P}}_1}}}{{{\text{P}}_1^0}} = {\chi _2}\,({\text{Raoult's}}\,{\text{law}})\) Relative lowering of vapour pressure is equal to the mole dfraction of solute. So option (2) is correct.
CHXII02:SOLUTIONS
319338
The vapour pressure of a solvent decreases by \(2.5 \mathrm{~mm} \mathrm{Hg}\) by adding a solute. What is the mole dfraction of solute? (Vapour pressure of pure solvent is \(250 \mathrm{~mm} \mathrm{Hg}\) )
1 0.88
2 0.01
3 0.1
4 0.99
Explanation:
\(\Delta {\text{P}} = 2.5\,{\text{mm Hg}},{\text{P}}_1^0 = 250\,{\text{mm Hg}}\) According to Raoult's law, \({{\text{x}}_2} = \frac{{\Delta {\text{P}}}}{{{\text{P}}_1^0}} = \frac{{2.5}}{{250}} = 0.01\)
319335
Vapour pressure of solution and of pure solvent are \(\mathrm{P}_{1}\) and \(\mathrm{P}_{1}^{0}\) respectively. If \(\dfrac{\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}\) is 0.15 , find the mole dfraction of solute.
1 0.66
2 0.85
3 0.15
4 0.33
Explanation:
\(\dfrac{\mathrm{P}_{1}^{0}-\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}=\mathrm{x}_{2}\) (mole dfraction of solute) \(\therefore \dfrac{\mathrm{P}_{1}^{0}}{\mathrm{P}_{1}^{0}}-\dfrac{\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}=\mathrm{x}_{2}\) \(\therefore \mathrm{x}_{2}=1-0.15=0.85\)
MHTCET - 2021
CHXII02:SOLUTIONS
319336
What is vapour pressure of a solution containing \(0.1 \mathrm{~mol}\) of non-volatile solute dissolved in \(16.2 \mathrm{~g}\) of water? \(\left(\mathrm{p}_{1}^{0}=32 \mathrm{~mm} \mathrm{Hg}\right)\)
319337
Assertion : The vapour pressure of a liquid decreases if some non-volatile solute is dissolved in it. Reason : The relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
The introduction of a non-volatile solute reduces the vapor pressure of a liquid solution by displacing solvent molecules at the surface, hindering their escape into the vapor phase. \({\text{RLVP}},{\text{ }}\frac{{\Delta {{\text{P}}_1}}}{{{\text{P}}_1^0}} = {\chi _2}\,({\text{Raoult's}}\,{\text{law}})\) Relative lowering of vapour pressure is equal to the mole dfraction of solute. So option (2) is correct.
CHXII02:SOLUTIONS
319338
The vapour pressure of a solvent decreases by \(2.5 \mathrm{~mm} \mathrm{Hg}\) by adding a solute. What is the mole dfraction of solute? (Vapour pressure of pure solvent is \(250 \mathrm{~mm} \mathrm{Hg}\) )
1 0.88
2 0.01
3 0.1
4 0.99
Explanation:
\(\Delta {\text{P}} = 2.5\,{\text{mm Hg}},{\text{P}}_1^0 = 250\,{\text{mm Hg}}\) According to Raoult's law, \({{\text{x}}_2} = \frac{{\Delta {\text{P}}}}{{{\text{P}}_1^0}} = \frac{{2.5}}{{250}} = 0.01\)
319335
Vapour pressure of solution and of pure solvent are \(\mathrm{P}_{1}\) and \(\mathrm{P}_{1}^{0}\) respectively. If \(\dfrac{\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}\) is 0.15 , find the mole dfraction of solute.
1 0.66
2 0.85
3 0.15
4 0.33
Explanation:
\(\dfrac{\mathrm{P}_{1}^{0}-\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}=\mathrm{x}_{2}\) (mole dfraction of solute) \(\therefore \dfrac{\mathrm{P}_{1}^{0}}{\mathrm{P}_{1}^{0}}-\dfrac{\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}=\mathrm{x}_{2}\) \(\therefore \mathrm{x}_{2}=1-0.15=0.85\)
MHTCET - 2021
CHXII02:SOLUTIONS
319336
What is vapour pressure of a solution containing \(0.1 \mathrm{~mol}\) of non-volatile solute dissolved in \(16.2 \mathrm{~g}\) of water? \(\left(\mathrm{p}_{1}^{0}=32 \mathrm{~mm} \mathrm{Hg}\right)\)
319337
Assertion : The vapour pressure of a liquid decreases if some non-volatile solute is dissolved in it. Reason : The relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
The introduction of a non-volatile solute reduces the vapor pressure of a liquid solution by displacing solvent molecules at the surface, hindering their escape into the vapor phase. \({\text{RLVP}},{\text{ }}\frac{{\Delta {{\text{P}}_1}}}{{{\text{P}}_1^0}} = {\chi _2}\,({\text{Raoult's}}\,{\text{law}})\) Relative lowering of vapour pressure is equal to the mole dfraction of solute. So option (2) is correct.
CHXII02:SOLUTIONS
319338
The vapour pressure of a solvent decreases by \(2.5 \mathrm{~mm} \mathrm{Hg}\) by adding a solute. What is the mole dfraction of solute? (Vapour pressure of pure solvent is \(250 \mathrm{~mm} \mathrm{Hg}\) )
1 0.88
2 0.01
3 0.1
4 0.99
Explanation:
\(\Delta {\text{P}} = 2.5\,{\text{mm Hg}},{\text{P}}_1^0 = 250\,{\text{mm Hg}}\) According to Raoult's law, \({{\text{x}}_2} = \frac{{\Delta {\text{P}}}}{{{\text{P}}_1^0}} = \frac{{2.5}}{{250}} = 0.01\)
319335
Vapour pressure of solution and of pure solvent are \(\mathrm{P}_{1}\) and \(\mathrm{P}_{1}^{0}\) respectively. If \(\dfrac{\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}\) is 0.15 , find the mole dfraction of solute.
1 0.66
2 0.85
3 0.15
4 0.33
Explanation:
\(\dfrac{\mathrm{P}_{1}^{0}-\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}=\mathrm{x}_{2}\) (mole dfraction of solute) \(\therefore \dfrac{\mathrm{P}_{1}^{0}}{\mathrm{P}_{1}^{0}}-\dfrac{\mathrm{P}_{1}}{\mathrm{P}_{1}^{0}}=\mathrm{x}_{2}\) \(\therefore \mathrm{x}_{2}=1-0.15=0.85\)
MHTCET - 2021
CHXII02:SOLUTIONS
319336
What is vapour pressure of a solution containing \(0.1 \mathrm{~mol}\) of non-volatile solute dissolved in \(16.2 \mathrm{~g}\) of water? \(\left(\mathrm{p}_{1}^{0}=32 \mathrm{~mm} \mathrm{Hg}\right)\)
319337
Assertion : The vapour pressure of a liquid decreases if some non-volatile solute is dissolved in it. Reason : The relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
The introduction of a non-volatile solute reduces the vapor pressure of a liquid solution by displacing solvent molecules at the surface, hindering their escape into the vapor phase. \({\text{RLVP}},{\text{ }}\frac{{\Delta {{\text{P}}_1}}}{{{\text{P}}_1^0}} = {\chi _2}\,({\text{Raoult's}}\,{\text{law}})\) Relative lowering of vapour pressure is equal to the mole dfraction of solute. So option (2) is correct.
CHXII02:SOLUTIONS
319338
The vapour pressure of a solvent decreases by \(2.5 \mathrm{~mm} \mathrm{Hg}\) by adding a solute. What is the mole dfraction of solute? (Vapour pressure of pure solvent is \(250 \mathrm{~mm} \mathrm{Hg}\) )
1 0.88
2 0.01
3 0.1
4 0.99
Explanation:
\(\Delta {\text{P}} = 2.5\,{\text{mm Hg}},{\text{P}}_1^0 = 250\,{\text{mm Hg}}\) According to Raoult's law, \({{\text{x}}_2} = \frac{{\Delta {\text{P}}}}{{{\text{P}}_1^0}} = \frac{{2.5}}{{250}} = 0.01\)
