277387
A 0.1 molal aqueous solution of a weak acid is $30 \%$ ionized. If $K_{f}$ for water is $1.86^{\circ} \mathrm{C} / \mathrm{m}$, the freezing point of the solution will be
277380
Assertion: Lowering of vapour pressure is directly proportional to osmotic pressure of the solution Reason: Osmotic pressure is a colligative property.
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
According to Van't Hoff equation for dilute solutions is- $\Pi=\frac{\mathrm{n}}{\mathrm{V}} \mathrm{RT}$ $\begin{array}{ll}\mathrm{V}=\frac{\mathrm{NM}}{\rho} & \begin{array}{l}\mathrm{N} \rightarrow \text { number of moles of solvent } \\ \mathrm{M} \rightarrow \text { molecular weight } \\ \rho \rightarrow \text { density } \\ \mathrm{V} \rightarrow \text { volume }\end{array} \\ \frac{\mathrm{n}}{\rho R \mathrm{M}}\end{array}$ $\begin{aligned} & \text { From Raoult's law } \\ & \frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=\frac{\mathrm{n}}{\mathrm{N}} \\ & \frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=\frac{\Pi \mathrm{M}}{\rho \mathrm{RT}} \\ & \mathrm{P}^{\mathrm{o}}-\mathrm{P}=\frac{\Pi \mathrm{M}}{\rho \mathrm{RT}} \times \mathrm{P}^{\mathrm{o}} \\ & \frac{\mathrm{MP}^{\mathrm{o}}}{\rho \mathrm{RT}} \text { factor is constant at constant temperature } \\ & \left(\mathrm{P}^{\mathrm{o}}-\mathrm{P}\right) \propto \Pi \end{aligned}$ lowering of vapour pressure $\propto$ osmotic pressure. The reason osmotic pressure is a colligative property is also true but not the correct explanation of the assertion.
AIIMS-2012
SOLUTIONS
277381
$12 \mathrm{~g}$ of urea is dissolved in 1 litre of water and $\mathbf{6 8 . 4} \mathrm{g}$ of sucrose is dissolved in 1 litre of water. The lowering of vapour pressure of first case is
1 equal to second
2 greater than second
3 less than second
4 double of that of second
Explanation:
Number of moles $=\frac{\text { Wt.of substance }}{\text { Molecular mass }}$ Molecular mass of urea $\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)$ $\begin{aligned} & =14+2+12+16+14+2 \\ & =16+12+16+16 \\ & =60 \end{aligned}$ Number of moles of urea in 1 liter of water $=\frac{12}{60}=\frac{1}{5}=0.2$ Molecular mass of sucrose $\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]$ $\begin{aligned} & =12 \times 12+1 \times 22+16 \times 11 \\ & =342 \end{aligned}$ Number of moles of sucrose in 1 liter of water $=\frac{68.4}{342}=\frac{1}{5}=0.2$ Urea and sucrose are non-electrolytic solutes and have same concentration. So, both witll have equal lowering of vapour pressure.
AIIMS-2012
SOLUTIONS
277382
The weight in grams of a non-volatile solute (mol. wt. 60) to be dissolved in $90 \mathrm{~g}$ of water to produce a relative lowering of vapour pressure of 0.02 is
1 4
2 8
3 6
4 10
Explanation:
Given that, Relative lowering of vapour pressure $\frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=0.02$ Molecular weight $\left(\mathrm{M}_{\mathrm{A}}\right)=60$ Weight of water $\left(\mathrm{W}_{\mathrm{A}}\right)=90$ gram $\frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=\mathrm{X}_{\mathrm{A}}$ $X_{A}=\frac{n_{A}}{n_{A}+n_{B}}=\frac{\frac{W_{A}}{M_{A}}}{\frac{W_{A}}{M_{A}}+\frac{W_{B}}{M_{B}}}$ $0.02=\frac{\frac{\mathrm{W}_{\mathrm{A}}}{60}}{\frac{\mathrm{W}_{\mathrm{A}}}{60}+\frac{90}{18}}$ $\frac{\mathrm{W}_{\mathrm{A}}}{60}+5=\frac{\mathrm{W}_{\mathrm{A}}}{0.02 \times 60}$ $\frac{\mathrm{W}_{\mathrm{A}}}{60}+5=\frac{\mathrm{W}_{\mathrm{A}}}{1.2}$ $5=\frac{\mathrm{W}_{\mathrm{A}}}{1.2}-\frac{\mathrm{W}_{\mathrm{A}}}{60}$ $\mathrm{W}_{\mathrm{A}}=6.122$ $\mathrm{W}_{\mathrm{A}} \approx 6$
277387
A 0.1 molal aqueous solution of a weak acid is $30 \%$ ionized. If $K_{f}$ for water is $1.86^{\circ} \mathrm{C} / \mathrm{m}$, the freezing point of the solution will be
277380
Assertion: Lowering of vapour pressure is directly proportional to osmotic pressure of the solution Reason: Osmotic pressure is a colligative property.
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
According to Van't Hoff equation for dilute solutions is- $\Pi=\frac{\mathrm{n}}{\mathrm{V}} \mathrm{RT}$ $\begin{array}{ll}\mathrm{V}=\frac{\mathrm{NM}}{\rho} & \begin{array}{l}\mathrm{N} \rightarrow \text { number of moles of solvent } \\ \mathrm{M} \rightarrow \text { molecular weight } \\ \rho \rightarrow \text { density } \\ \mathrm{V} \rightarrow \text { volume }\end{array} \\ \frac{\mathrm{n}}{\rho R \mathrm{M}}\end{array}$ $\begin{aligned} & \text { From Raoult's law } \\ & \frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=\frac{\mathrm{n}}{\mathrm{N}} \\ & \frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=\frac{\Pi \mathrm{M}}{\rho \mathrm{RT}} \\ & \mathrm{P}^{\mathrm{o}}-\mathrm{P}=\frac{\Pi \mathrm{M}}{\rho \mathrm{RT}} \times \mathrm{P}^{\mathrm{o}} \\ & \frac{\mathrm{MP}^{\mathrm{o}}}{\rho \mathrm{RT}} \text { factor is constant at constant temperature } \\ & \left(\mathrm{P}^{\mathrm{o}}-\mathrm{P}\right) \propto \Pi \end{aligned}$ lowering of vapour pressure $\propto$ osmotic pressure. The reason osmotic pressure is a colligative property is also true but not the correct explanation of the assertion.
AIIMS-2012
SOLUTIONS
277381
$12 \mathrm{~g}$ of urea is dissolved in 1 litre of water and $\mathbf{6 8 . 4} \mathrm{g}$ of sucrose is dissolved in 1 litre of water. The lowering of vapour pressure of first case is
1 equal to second
2 greater than second
3 less than second
4 double of that of second
Explanation:
Number of moles $=\frac{\text { Wt.of substance }}{\text { Molecular mass }}$ Molecular mass of urea $\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)$ $\begin{aligned} & =14+2+12+16+14+2 \\ & =16+12+16+16 \\ & =60 \end{aligned}$ Number of moles of urea in 1 liter of water $=\frac{12}{60}=\frac{1}{5}=0.2$ Molecular mass of sucrose $\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]$ $\begin{aligned} & =12 \times 12+1 \times 22+16 \times 11 \\ & =342 \end{aligned}$ Number of moles of sucrose in 1 liter of water $=\frac{68.4}{342}=\frac{1}{5}=0.2$ Urea and sucrose are non-electrolytic solutes and have same concentration. So, both witll have equal lowering of vapour pressure.
AIIMS-2012
SOLUTIONS
277382
The weight in grams of a non-volatile solute (mol. wt. 60) to be dissolved in $90 \mathrm{~g}$ of water to produce a relative lowering of vapour pressure of 0.02 is
1 4
2 8
3 6
4 10
Explanation:
Given that, Relative lowering of vapour pressure $\frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=0.02$ Molecular weight $\left(\mathrm{M}_{\mathrm{A}}\right)=60$ Weight of water $\left(\mathrm{W}_{\mathrm{A}}\right)=90$ gram $\frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=\mathrm{X}_{\mathrm{A}}$ $X_{A}=\frac{n_{A}}{n_{A}+n_{B}}=\frac{\frac{W_{A}}{M_{A}}}{\frac{W_{A}}{M_{A}}+\frac{W_{B}}{M_{B}}}$ $0.02=\frac{\frac{\mathrm{W}_{\mathrm{A}}}{60}}{\frac{\mathrm{W}_{\mathrm{A}}}{60}+\frac{90}{18}}$ $\frac{\mathrm{W}_{\mathrm{A}}}{60}+5=\frac{\mathrm{W}_{\mathrm{A}}}{0.02 \times 60}$ $\frac{\mathrm{W}_{\mathrm{A}}}{60}+5=\frac{\mathrm{W}_{\mathrm{A}}}{1.2}$ $5=\frac{\mathrm{W}_{\mathrm{A}}}{1.2}-\frac{\mathrm{W}_{\mathrm{A}}}{60}$ $\mathrm{W}_{\mathrm{A}}=6.122$ $\mathrm{W}_{\mathrm{A}} \approx 6$
277387
A 0.1 molal aqueous solution of a weak acid is $30 \%$ ionized. If $K_{f}$ for water is $1.86^{\circ} \mathrm{C} / \mathrm{m}$, the freezing point of the solution will be
277380
Assertion: Lowering of vapour pressure is directly proportional to osmotic pressure of the solution Reason: Osmotic pressure is a colligative property.
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
According to Van't Hoff equation for dilute solutions is- $\Pi=\frac{\mathrm{n}}{\mathrm{V}} \mathrm{RT}$ $\begin{array}{ll}\mathrm{V}=\frac{\mathrm{NM}}{\rho} & \begin{array}{l}\mathrm{N} \rightarrow \text { number of moles of solvent } \\ \mathrm{M} \rightarrow \text { molecular weight } \\ \rho \rightarrow \text { density } \\ \mathrm{V} \rightarrow \text { volume }\end{array} \\ \frac{\mathrm{n}}{\rho R \mathrm{M}}\end{array}$ $\begin{aligned} & \text { From Raoult's law } \\ & \frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=\frac{\mathrm{n}}{\mathrm{N}} \\ & \frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=\frac{\Pi \mathrm{M}}{\rho \mathrm{RT}} \\ & \mathrm{P}^{\mathrm{o}}-\mathrm{P}=\frac{\Pi \mathrm{M}}{\rho \mathrm{RT}} \times \mathrm{P}^{\mathrm{o}} \\ & \frac{\mathrm{MP}^{\mathrm{o}}}{\rho \mathrm{RT}} \text { factor is constant at constant temperature } \\ & \left(\mathrm{P}^{\mathrm{o}}-\mathrm{P}\right) \propto \Pi \end{aligned}$ lowering of vapour pressure $\propto$ osmotic pressure. The reason osmotic pressure is a colligative property is also true but not the correct explanation of the assertion.
AIIMS-2012
SOLUTIONS
277381
$12 \mathrm{~g}$ of urea is dissolved in 1 litre of water and $\mathbf{6 8 . 4} \mathrm{g}$ of sucrose is dissolved in 1 litre of water. The lowering of vapour pressure of first case is
1 equal to second
2 greater than second
3 less than second
4 double of that of second
Explanation:
Number of moles $=\frac{\text { Wt.of substance }}{\text { Molecular mass }}$ Molecular mass of urea $\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)$ $\begin{aligned} & =14+2+12+16+14+2 \\ & =16+12+16+16 \\ & =60 \end{aligned}$ Number of moles of urea in 1 liter of water $=\frac{12}{60}=\frac{1}{5}=0.2$ Molecular mass of sucrose $\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]$ $\begin{aligned} & =12 \times 12+1 \times 22+16 \times 11 \\ & =342 \end{aligned}$ Number of moles of sucrose in 1 liter of water $=\frac{68.4}{342}=\frac{1}{5}=0.2$ Urea and sucrose are non-electrolytic solutes and have same concentration. So, both witll have equal lowering of vapour pressure.
AIIMS-2012
SOLUTIONS
277382
The weight in grams of a non-volatile solute (mol. wt. 60) to be dissolved in $90 \mathrm{~g}$ of water to produce a relative lowering of vapour pressure of 0.02 is
1 4
2 8
3 6
4 10
Explanation:
Given that, Relative lowering of vapour pressure $\frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=0.02$ Molecular weight $\left(\mathrm{M}_{\mathrm{A}}\right)=60$ Weight of water $\left(\mathrm{W}_{\mathrm{A}}\right)=90$ gram $\frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=\mathrm{X}_{\mathrm{A}}$ $X_{A}=\frac{n_{A}}{n_{A}+n_{B}}=\frac{\frac{W_{A}}{M_{A}}}{\frac{W_{A}}{M_{A}}+\frac{W_{B}}{M_{B}}}$ $0.02=\frac{\frac{\mathrm{W}_{\mathrm{A}}}{60}}{\frac{\mathrm{W}_{\mathrm{A}}}{60}+\frac{90}{18}}$ $\frac{\mathrm{W}_{\mathrm{A}}}{60}+5=\frac{\mathrm{W}_{\mathrm{A}}}{0.02 \times 60}$ $\frac{\mathrm{W}_{\mathrm{A}}}{60}+5=\frac{\mathrm{W}_{\mathrm{A}}}{1.2}$ $5=\frac{\mathrm{W}_{\mathrm{A}}}{1.2}-\frac{\mathrm{W}_{\mathrm{A}}}{60}$ $\mathrm{W}_{\mathrm{A}}=6.122$ $\mathrm{W}_{\mathrm{A}} \approx 6$
NEET Test Series from KOTA - 10 Papers In MS WORD
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SOLUTIONS
277387
A 0.1 molal aqueous solution of a weak acid is $30 \%$ ionized. If $K_{f}$ for water is $1.86^{\circ} \mathrm{C} / \mathrm{m}$, the freezing point of the solution will be
277380
Assertion: Lowering of vapour pressure is directly proportional to osmotic pressure of the solution Reason: Osmotic pressure is a colligative property.
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
According to Van't Hoff equation for dilute solutions is- $\Pi=\frac{\mathrm{n}}{\mathrm{V}} \mathrm{RT}$ $\begin{array}{ll}\mathrm{V}=\frac{\mathrm{NM}}{\rho} & \begin{array}{l}\mathrm{N} \rightarrow \text { number of moles of solvent } \\ \mathrm{M} \rightarrow \text { molecular weight } \\ \rho \rightarrow \text { density } \\ \mathrm{V} \rightarrow \text { volume }\end{array} \\ \frac{\mathrm{n}}{\rho R \mathrm{M}}\end{array}$ $\begin{aligned} & \text { From Raoult's law } \\ & \frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=\frac{\mathrm{n}}{\mathrm{N}} \\ & \frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=\frac{\Pi \mathrm{M}}{\rho \mathrm{RT}} \\ & \mathrm{P}^{\mathrm{o}}-\mathrm{P}=\frac{\Pi \mathrm{M}}{\rho \mathrm{RT}} \times \mathrm{P}^{\mathrm{o}} \\ & \frac{\mathrm{MP}^{\mathrm{o}}}{\rho \mathrm{RT}} \text { factor is constant at constant temperature } \\ & \left(\mathrm{P}^{\mathrm{o}}-\mathrm{P}\right) \propto \Pi \end{aligned}$ lowering of vapour pressure $\propto$ osmotic pressure. The reason osmotic pressure is a colligative property is also true but not the correct explanation of the assertion.
AIIMS-2012
SOLUTIONS
277381
$12 \mathrm{~g}$ of urea is dissolved in 1 litre of water and $\mathbf{6 8 . 4} \mathrm{g}$ of sucrose is dissolved in 1 litre of water. The lowering of vapour pressure of first case is
1 equal to second
2 greater than second
3 less than second
4 double of that of second
Explanation:
Number of moles $=\frac{\text { Wt.of substance }}{\text { Molecular mass }}$ Molecular mass of urea $\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)$ $\begin{aligned} & =14+2+12+16+14+2 \\ & =16+12+16+16 \\ & =60 \end{aligned}$ Number of moles of urea in 1 liter of water $=\frac{12}{60}=\frac{1}{5}=0.2$ Molecular mass of sucrose $\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]$ $\begin{aligned} & =12 \times 12+1 \times 22+16 \times 11 \\ & =342 \end{aligned}$ Number of moles of sucrose in 1 liter of water $=\frac{68.4}{342}=\frac{1}{5}=0.2$ Urea and sucrose are non-electrolytic solutes and have same concentration. So, both witll have equal lowering of vapour pressure.
AIIMS-2012
SOLUTIONS
277382
The weight in grams of a non-volatile solute (mol. wt. 60) to be dissolved in $90 \mathrm{~g}$ of water to produce a relative lowering of vapour pressure of 0.02 is
1 4
2 8
3 6
4 10
Explanation:
Given that, Relative lowering of vapour pressure $\frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=0.02$ Molecular weight $\left(\mathrm{M}_{\mathrm{A}}\right)=60$ Weight of water $\left(\mathrm{W}_{\mathrm{A}}\right)=90$ gram $\frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=\mathrm{X}_{\mathrm{A}}$ $X_{A}=\frac{n_{A}}{n_{A}+n_{B}}=\frac{\frac{W_{A}}{M_{A}}}{\frac{W_{A}}{M_{A}}+\frac{W_{B}}{M_{B}}}$ $0.02=\frac{\frac{\mathrm{W}_{\mathrm{A}}}{60}}{\frac{\mathrm{W}_{\mathrm{A}}}{60}+\frac{90}{18}}$ $\frac{\mathrm{W}_{\mathrm{A}}}{60}+5=\frac{\mathrm{W}_{\mathrm{A}}}{0.02 \times 60}$ $\frac{\mathrm{W}_{\mathrm{A}}}{60}+5=\frac{\mathrm{W}_{\mathrm{A}}}{1.2}$ $5=\frac{\mathrm{W}_{\mathrm{A}}}{1.2}-\frac{\mathrm{W}_{\mathrm{A}}}{60}$ $\mathrm{W}_{\mathrm{A}}=6.122$ $\mathrm{W}_{\mathrm{A}} \approx 6$
