277520
A solution containing $6.0 \mathrm{~g}$ of urea is isotonic with a solution containing $10 \mathrm{~g}$ of a nonelectrolytic solute $X$. The molar mass of $X$ (in $g$ mol $^{-1}$ ) is
1 50.0
2 100
3 75.0
4 68.0
Explanation:
Weight of area $=6.0 \mathrm{gm}$ Molecular weight of area $=60 \mathrm{~g} \mathrm{~mol}^{-1}$ Weight of solute $\times=10 \mathrm{~g}$ For, isotonic solution $\lambda_{1}=\lambda_{2}$ Or $\mathrm{C}_{1}=\mathrm{C}_{2}$ (conc. In mol/lit) Where, $\mathrm{C}=\frac{\text { weight }}{\text { molecular weight }}$ So, $\quad \frac{6.0}{60}=\frac{10}{\mathrm{~m}_{\mathrm{w}}}$ $\begin{aligned} & \mathrm{m}_{\mathrm{w}}=\frac{60 \times 10}{6.0} \\ & \mathrm{~m}_{\mathrm{w}}=100 \mathrm{~g} \mathrm{~mol}^{-1} \end{aligned}$
AP-EAMCET-04.07.2022 Shift-II
SOLUTIONS
277521
$x \%(w / v)$ solution of urea is isotonic with $4 \%$ $(w / v)$ solution of a non-volatile solute of molar mass $120 \mathrm{~g} \mathrm{~mol}^{-1}$. The value of $x$ is
277523
Solutions $A, B, C$ and $D$ are $0.1 \mathrm{M}$ glucose, 0.05 $\mathrm{M} \mathrm{NaCl}, \mathbf{0 . 0 5} \mathrm{M}^{\mathrm{BaCl}_{2}}$ and $0.1 \mathrm{M} \mathrm{AlCl}_{3}$ respectively. Which one of the following pairs is isotonic?
1 $\mathrm{B}$ and $\mathrm{C}$
2 A and B
3 A and D
4 A and C
Explanation:
The solutions which have same osmotic pressure is known as isotonic solution. $\Pi=\mathrm{i} \text { C.R.T }$ Where $-\Pi=$ Osmotic pressure $\mathrm{C}=$ Concentration of solution $\mathrm{R}=$ Gas constant $\mathrm{T}=$ Temperature $\mathrm{i}=$ Van't Hoff factor. For solution $\mathrm{A}$ - $\prod_{\mathrm{A}}=1 \times 0.1 \mathrm{RT}=0.1 \mathrm{RT}$ For solution $\mathrm{B}$ - For solution $\mathrm{C}$ - $\prod_{\mathrm{B}}=2 \times 0.05 \mathrm{RT}=0.1 \mathrm{RT}$ For solution D - $\prod_{\mathrm{C}}=3 \times 0.05 \mathrm{RT}=0.15 \mathrm{RT}$ $\prod_{\mathrm{D}}=4 \times 0.1 \mathrm{RT}=0.4 \mathrm{RT}$ Thus, solution A and B have the same osmotic pressure therefore isotonic solution.
JIPMER-2017
SOLUTIONS
277528
A $25 \%$ solution of cane-sugar (mol mass $=342$ $\mathrm{g} \mathrm{mol}^{-1}$ ) is isotonic with $5 \%$ solution of a substance $A$. Then find the molecular weight of A?
1 $6.84 \mathrm{~g} \mathrm{~mol}^{-1}$
2 $68.4 \mathrm{~g} \mathrm{~mol}^{-1}$
3 $25 \mathrm{~g} \mathrm{~mol}^{-1}$
4 $684 \mathrm{~g} \mathrm{~mol}^{-1}$
Explanation:
Given, $25 \%$ solution of cane sugar means $25 \mathrm{~g}$ of solute present in $100 \mathrm{~mL}$ of solution. $\therefore$ For isotonic solutions, $\prod_{1}=\prod_{2}$ or $\quad \mathrm{C}_{1}=\mathrm{C}_{2}$ $\begin{aligned} & \frac{25}{342 \times 0.1}=\frac{5}{\mathrm{M}_{2} \times 0.1} \\ & \mathrm{M}_{2}=\frac{342}{5}=68.4 \mathrm{~g} / \mathrm{mol} \end{aligned}$
AP-EAMCET 25-08-2021 Shift - I
SOLUTIONS
277530
Which of the following pairs of solutions is expected to be isotonic at same temperature?
1 $0.1 \mathrm{M}$ urea and $0.1 \mathrm{M} \mathrm{NaCl}$
2 $0.1 \mathrm{M}$ glucose and $0.2 \mathrm{M} \mathrm{NaCl}$
3 $0.1 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}$
4 $0.1 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{KNO}_{3}$
Explanation:
Two solutions of same osmotic pressure (П) are isotonic solutions. $\begin{aligned} & \prod_{1}(0.1 \mathrm{M} \mathrm{NaCl})=0.1 \times \mathrm{RT} \times 2 \\ & \prod_{2}\left(0.1 \mathrm{M} \mathrm{KNO}_{3}\right)=0.1 \times \mathrm{RT} \times 2 \end{aligned}$ So, $\Pi_{1}(0.1 \mathrm{M} \mathrm{NaCl})=\prod_{2}\left(0.1 \mathrm{M} \mathrm{KNO}_{3}\right)$ at same temperature.
277520
A solution containing $6.0 \mathrm{~g}$ of urea is isotonic with a solution containing $10 \mathrm{~g}$ of a nonelectrolytic solute $X$. The molar mass of $X$ (in $g$ mol $^{-1}$ ) is
1 50.0
2 100
3 75.0
4 68.0
Explanation:
Weight of area $=6.0 \mathrm{gm}$ Molecular weight of area $=60 \mathrm{~g} \mathrm{~mol}^{-1}$ Weight of solute $\times=10 \mathrm{~g}$ For, isotonic solution $\lambda_{1}=\lambda_{2}$ Or $\mathrm{C}_{1}=\mathrm{C}_{2}$ (conc. In mol/lit) Where, $\mathrm{C}=\frac{\text { weight }}{\text { molecular weight }}$ So, $\quad \frac{6.0}{60}=\frac{10}{\mathrm{~m}_{\mathrm{w}}}$ $\begin{aligned} & \mathrm{m}_{\mathrm{w}}=\frac{60 \times 10}{6.0} \\ & \mathrm{~m}_{\mathrm{w}}=100 \mathrm{~g} \mathrm{~mol}^{-1} \end{aligned}$
AP-EAMCET-04.07.2022 Shift-II
SOLUTIONS
277521
$x \%(w / v)$ solution of urea is isotonic with $4 \%$ $(w / v)$ solution of a non-volatile solute of molar mass $120 \mathrm{~g} \mathrm{~mol}^{-1}$. The value of $x$ is
277523
Solutions $A, B, C$ and $D$ are $0.1 \mathrm{M}$ glucose, 0.05 $\mathrm{M} \mathrm{NaCl}, \mathbf{0 . 0 5} \mathrm{M}^{\mathrm{BaCl}_{2}}$ and $0.1 \mathrm{M} \mathrm{AlCl}_{3}$ respectively. Which one of the following pairs is isotonic?
1 $\mathrm{B}$ and $\mathrm{C}$
2 A and B
3 A and D
4 A and C
Explanation:
The solutions which have same osmotic pressure is known as isotonic solution. $\Pi=\mathrm{i} \text { C.R.T }$ Where $-\Pi=$ Osmotic pressure $\mathrm{C}=$ Concentration of solution $\mathrm{R}=$ Gas constant $\mathrm{T}=$ Temperature $\mathrm{i}=$ Van't Hoff factor. For solution $\mathrm{A}$ - $\prod_{\mathrm{A}}=1 \times 0.1 \mathrm{RT}=0.1 \mathrm{RT}$ For solution $\mathrm{B}$ - For solution $\mathrm{C}$ - $\prod_{\mathrm{B}}=2 \times 0.05 \mathrm{RT}=0.1 \mathrm{RT}$ For solution D - $\prod_{\mathrm{C}}=3 \times 0.05 \mathrm{RT}=0.15 \mathrm{RT}$ $\prod_{\mathrm{D}}=4 \times 0.1 \mathrm{RT}=0.4 \mathrm{RT}$ Thus, solution A and B have the same osmotic pressure therefore isotonic solution.
JIPMER-2017
SOLUTIONS
277528
A $25 \%$ solution of cane-sugar (mol mass $=342$ $\mathrm{g} \mathrm{mol}^{-1}$ ) is isotonic with $5 \%$ solution of a substance $A$. Then find the molecular weight of A?
1 $6.84 \mathrm{~g} \mathrm{~mol}^{-1}$
2 $68.4 \mathrm{~g} \mathrm{~mol}^{-1}$
3 $25 \mathrm{~g} \mathrm{~mol}^{-1}$
4 $684 \mathrm{~g} \mathrm{~mol}^{-1}$
Explanation:
Given, $25 \%$ solution of cane sugar means $25 \mathrm{~g}$ of solute present in $100 \mathrm{~mL}$ of solution. $\therefore$ For isotonic solutions, $\prod_{1}=\prod_{2}$ or $\quad \mathrm{C}_{1}=\mathrm{C}_{2}$ $\begin{aligned} & \frac{25}{342 \times 0.1}=\frac{5}{\mathrm{M}_{2} \times 0.1} \\ & \mathrm{M}_{2}=\frac{342}{5}=68.4 \mathrm{~g} / \mathrm{mol} \end{aligned}$
AP-EAMCET 25-08-2021 Shift - I
SOLUTIONS
277530
Which of the following pairs of solutions is expected to be isotonic at same temperature?
1 $0.1 \mathrm{M}$ urea and $0.1 \mathrm{M} \mathrm{NaCl}$
2 $0.1 \mathrm{M}$ glucose and $0.2 \mathrm{M} \mathrm{NaCl}$
3 $0.1 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}$
4 $0.1 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{KNO}_{3}$
Explanation:
Two solutions of same osmotic pressure (П) are isotonic solutions. $\begin{aligned} & \prod_{1}(0.1 \mathrm{M} \mathrm{NaCl})=0.1 \times \mathrm{RT} \times 2 \\ & \prod_{2}\left(0.1 \mathrm{M} \mathrm{KNO}_{3}\right)=0.1 \times \mathrm{RT} \times 2 \end{aligned}$ So, $\Pi_{1}(0.1 \mathrm{M} \mathrm{NaCl})=\prod_{2}\left(0.1 \mathrm{M} \mathrm{KNO}_{3}\right)$ at same temperature.
277520
A solution containing $6.0 \mathrm{~g}$ of urea is isotonic with a solution containing $10 \mathrm{~g}$ of a nonelectrolytic solute $X$. The molar mass of $X$ (in $g$ mol $^{-1}$ ) is
1 50.0
2 100
3 75.0
4 68.0
Explanation:
Weight of area $=6.0 \mathrm{gm}$ Molecular weight of area $=60 \mathrm{~g} \mathrm{~mol}^{-1}$ Weight of solute $\times=10 \mathrm{~g}$ For, isotonic solution $\lambda_{1}=\lambda_{2}$ Or $\mathrm{C}_{1}=\mathrm{C}_{2}$ (conc. In mol/lit) Where, $\mathrm{C}=\frac{\text { weight }}{\text { molecular weight }}$ So, $\quad \frac{6.0}{60}=\frac{10}{\mathrm{~m}_{\mathrm{w}}}$ $\begin{aligned} & \mathrm{m}_{\mathrm{w}}=\frac{60 \times 10}{6.0} \\ & \mathrm{~m}_{\mathrm{w}}=100 \mathrm{~g} \mathrm{~mol}^{-1} \end{aligned}$
AP-EAMCET-04.07.2022 Shift-II
SOLUTIONS
277521
$x \%(w / v)$ solution of urea is isotonic with $4 \%$ $(w / v)$ solution of a non-volatile solute of molar mass $120 \mathrm{~g} \mathrm{~mol}^{-1}$. The value of $x$ is
277523
Solutions $A, B, C$ and $D$ are $0.1 \mathrm{M}$ glucose, 0.05 $\mathrm{M} \mathrm{NaCl}, \mathbf{0 . 0 5} \mathrm{M}^{\mathrm{BaCl}_{2}}$ and $0.1 \mathrm{M} \mathrm{AlCl}_{3}$ respectively. Which one of the following pairs is isotonic?
1 $\mathrm{B}$ and $\mathrm{C}$
2 A and B
3 A and D
4 A and C
Explanation:
The solutions which have same osmotic pressure is known as isotonic solution. $\Pi=\mathrm{i} \text { C.R.T }$ Where $-\Pi=$ Osmotic pressure $\mathrm{C}=$ Concentration of solution $\mathrm{R}=$ Gas constant $\mathrm{T}=$ Temperature $\mathrm{i}=$ Van't Hoff factor. For solution $\mathrm{A}$ - $\prod_{\mathrm{A}}=1 \times 0.1 \mathrm{RT}=0.1 \mathrm{RT}$ For solution $\mathrm{B}$ - For solution $\mathrm{C}$ - $\prod_{\mathrm{B}}=2 \times 0.05 \mathrm{RT}=0.1 \mathrm{RT}$ For solution D - $\prod_{\mathrm{C}}=3 \times 0.05 \mathrm{RT}=0.15 \mathrm{RT}$ $\prod_{\mathrm{D}}=4 \times 0.1 \mathrm{RT}=0.4 \mathrm{RT}$ Thus, solution A and B have the same osmotic pressure therefore isotonic solution.
JIPMER-2017
SOLUTIONS
277528
A $25 \%$ solution of cane-sugar (mol mass $=342$ $\mathrm{g} \mathrm{mol}^{-1}$ ) is isotonic with $5 \%$ solution of a substance $A$. Then find the molecular weight of A?
1 $6.84 \mathrm{~g} \mathrm{~mol}^{-1}$
2 $68.4 \mathrm{~g} \mathrm{~mol}^{-1}$
3 $25 \mathrm{~g} \mathrm{~mol}^{-1}$
4 $684 \mathrm{~g} \mathrm{~mol}^{-1}$
Explanation:
Given, $25 \%$ solution of cane sugar means $25 \mathrm{~g}$ of solute present in $100 \mathrm{~mL}$ of solution. $\therefore$ For isotonic solutions, $\prod_{1}=\prod_{2}$ or $\quad \mathrm{C}_{1}=\mathrm{C}_{2}$ $\begin{aligned} & \frac{25}{342 \times 0.1}=\frac{5}{\mathrm{M}_{2} \times 0.1} \\ & \mathrm{M}_{2}=\frac{342}{5}=68.4 \mathrm{~g} / \mathrm{mol} \end{aligned}$
AP-EAMCET 25-08-2021 Shift - I
SOLUTIONS
277530
Which of the following pairs of solutions is expected to be isotonic at same temperature?
1 $0.1 \mathrm{M}$ urea and $0.1 \mathrm{M} \mathrm{NaCl}$
2 $0.1 \mathrm{M}$ glucose and $0.2 \mathrm{M} \mathrm{NaCl}$
3 $0.1 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}$
4 $0.1 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{KNO}_{3}$
Explanation:
Two solutions of same osmotic pressure (П) are isotonic solutions. $\begin{aligned} & \prod_{1}(0.1 \mathrm{M} \mathrm{NaCl})=0.1 \times \mathrm{RT} \times 2 \\ & \prod_{2}\left(0.1 \mathrm{M} \mathrm{KNO}_{3}\right)=0.1 \times \mathrm{RT} \times 2 \end{aligned}$ So, $\Pi_{1}(0.1 \mathrm{M} \mathrm{NaCl})=\prod_{2}\left(0.1 \mathrm{M} \mathrm{KNO}_{3}\right)$ at same temperature.
277520
A solution containing $6.0 \mathrm{~g}$ of urea is isotonic with a solution containing $10 \mathrm{~g}$ of a nonelectrolytic solute $X$. The molar mass of $X$ (in $g$ mol $^{-1}$ ) is
1 50.0
2 100
3 75.0
4 68.0
Explanation:
Weight of area $=6.0 \mathrm{gm}$ Molecular weight of area $=60 \mathrm{~g} \mathrm{~mol}^{-1}$ Weight of solute $\times=10 \mathrm{~g}$ For, isotonic solution $\lambda_{1}=\lambda_{2}$ Or $\mathrm{C}_{1}=\mathrm{C}_{2}$ (conc. In mol/lit) Where, $\mathrm{C}=\frac{\text { weight }}{\text { molecular weight }}$ So, $\quad \frac{6.0}{60}=\frac{10}{\mathrm{~m}_{\mathrm{w}}}$ $\begin{aligned} & \mathrm{m}_{\mathrm{w}}=\frac{60 \times 10}{6.0} \\ & \mathrm{~m}_{\mathrm{w}}=100 \mathrm{~g} \mathrm{~mol}^{-1} \end{aligned}$
AP-EAMCET-04.07.2022 Shift-II
SOLUTIONS
277521
$x \%(w / v)$ solution of urea is isotonic with $4 \%$ $(w / v)$ solution of a non-volatile solute of molar mass $120 \mathrm{~g} \mathrm{~mol}^{-1}$. The value of $x$ is
277523
Solutions $A, B, C$ and $D$ are $0.1 \mathrm{M}$ glucose, 0.05 $\mathrm{M} \mathrm{NaCl}, \mathbf{0 . 0 5} \mathrm{M}^{\mathrm{BaCl}_{2}}$ and $0.1 \mathrm{M} \mathrm{AlCl}_{3}$ respectively. Which one of the following pairs is isotonic?
1 $\mathrm{B}$ and $\mathrm{C}$
2 A and B
3 A and D
4 A and C
Explanation:
The solutions which have same osmotic pressure is known as isotonic solution. $\Pi=\mathrm{i} \text { C.R.T }$ Where $-\Pi=$ Osmotic pressure $\mathrm{C}=$ Concentration of solution $\mathrm{R}=$ Gas constant $\mathrm{T}=$ Temperature $\mathrm{i}=$ Van't Hoff factor. For solution $\mathrm{A}$ - $\prod_{\mathrm{A}}=1 \times 0.1 \mathrm{RT}=0.1 \mathrm{RT}$ For solution $\mathrm{B}$ - For solution $\mathrm{C}$ - $\prod_{\mathrm{B}}=2 \times 0.05 \mathrm{RT}=0.1 \mathrm{RT}$ For solution D - $\prod_{\mathrm{C}}=3 \times 0.05 \mathrm{RT}=0.15 \mathrm{RT}$ $\prod_{\mathrm{D}}=4 \times 0.1 \mathrm{RT}=0.4 \mathrm{RT}$ Thus, solution A and B have the same osmotic pressure therefore isotonic solution.
JIPMER-2017
SOLUTIONS
277528
A $25 \%$ solution of cane-sugar (mol mass $=342$ $\mathrm{g} \mathrm{mol}^{-1}$ ) is isotonic with $5 \%$ solution of a substance $A$. Then find the molecular weight of A?
1 $6.84 \mathrm{~g} \mathrm{~mol}^{-1}$
2 $68.4 \mathrm{~g} \mathrm{~mol}^{-1}$
3 $25 \mathrm{~g} \mathrm{~mol}^{-1}$
4 $684 \mathrm{~g} \mathrm{~mol}^{-1}$
Explanation:
Given, $25 \%$ solution of cane sugar means $25 \mathrm{~g}$ of solute present in $100 \mathrm{~mL}$ of solution. $\therefore$ For isotonic solutions, $\prod_{1}=\prod_{2}$ or $\quad \mathrm{C}_{1}=\mathrm{C}_{2}$ $\begin{aligned} & \frac{25}{342 \times 0.1}=\frac{5}{\mathrm{M}_{2} \times 0.1} \\ & \mathrm{M}_{2}=\frac{342}{5}=68.4 \mathrm{~g} / \mathrm{mol} \end{aligned}$
AP-EAMCET 25-08-2021 Shift - I
SOLUTIONS
277530
Which of the following pairs of solutions is expected to be isotonic at same temperature?
1 $0.1 \mathrm{M}$ urea and $0.1 \mathrm{M} \mathrm{NaCl}$
2 $0.1 \mathrm{M}$ glucose and $0.2 \mathrm{M} \mathrm{NaCl}$
3 $0.1 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}$
4 $0.1 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{KNO}_{3}$
Explanation:
Two solutions of same osmotic pressure (П) are isotonic solutions. $\begin{aligned} & \prod_{1}(0.1 \mathrm{M} \mathrm{NaCl})=0.1 \times \mathrm{RT} \times 2 \\ & \prod_{2}\left(0.1 \mathrm{M} \mathrm{KNO}_{3}\right)=0.1 \times \mathrm{RT} \times 2 \end{aligned}$ So, $\Pi_{1}(0.1 \mathrm{M} \mathrm{NaCl})=\prod_{2}\left(0.1 \mathrm{M} \mathrm{KNO}_{3}\right)$ at same temperature.
277520
A solution containing $6.0 \mathrm{~g}$ of urea is isotonic with a solution containing $10 \mathrm{~g}$ of a nonelectrolytic solute $X$. The molar mass of $X$ (in $g$ mol $^{-1}$ ) is
1 50.0
2 100
3 75.0
4 68.0
Explanation:
Weight of area $=6.0 \mathrm{gm}$ Molecular weight of area $=60 \mathrm{~g} \mathrm{~mol}^{-1}$ Weight of solute $\times=10 \mathrm{~g}$ For, isotonic solution $\lambda_{1}=\lambda_{2}$ Or $\mathrm{C}_{1}=\mathrm{C}_{2}$ (conc. In mol/lit) Where, $\mathrm{C}=\frac{\text { weight }}{\text { molecular weight }}$ So, $\quad \frac{6.0}{60}=\frac{10}{\mathrm{~m}_{\mathrm{w}}}$ $\begin{aligned} & \mathrm{m}_{\mathrm{w}}=\frac{60 \times 10}{6.0} \\ & \mathrm{~m}_{\mathrm{w}}=100 \mathrm{~g} \mathrm{~mol}^{-1} \end{aligned}$
AP-EAMCET-04.07.2022 Shift-II
SOLUTIONS
277521
$x \%(w / v)$ solution of urea is isotonic with $4 \%$ $(w / v)$ solution of a non-volatile solute of molar mass $120 \mathrm{~g} \mathrm{~mol}^{-1}$. The value of $x$ is
277523
Solutions $A, B, C$ and $D$ are $0.1 \mathrm{M}$ glucose, 0.05 $\mathrm{M} \mathrm{NaCl}, \mathbf{0 . 0 5} \mathrm{M}^{\mathrm{BaCl}_{2}}$ and $0.1 \mathrm{M} \mathrm{AlCl}_{3}$ respectively. Which one of the following pairs is isotonic?
1 $\mathrm{B}$ and $\mathrm{C}$
2 A and B
3 A and D
4 A and C
Explanation:
The solutions which have same osmotic pressure is known as isotonic solution. $\Pi=\mathrm{i} \text { C.R.T }$ Where $-\Pi=$ Osmotic pressure $\mathrm{C}=$ Concentration of solution $\mathrm{R}=$ Gas constant $\mathrm{T}=$ Temperature $\mathrm{i}=$ Van't Hoff factor. For solution $\mathrm{A}$ - $\prod_{\mathrm{A}}=1 \times 0.1 \mathrm{RT}=0.1 \mathrm{RT}$ For solution $\mathrm{B}$ - For solution $\mathrm{C}$ - $\prod_{\mathrm{B}}=2 \times 0.05 \mathrm{RT}=0.1 \mathrm{RT}$ For solution D - $\prod_{\mathrm{C}}=3 \times 0.05 \mathrm{RT}=0.15 \mathrm{RT}$ $\prod_{\mathrm{D}}=4 \times 0.1 \mathrm{RT}=0.4 \mathrm{RT}$ Thus, solution A and B have the same osmotic pressure therefore isotonic solution.
JIPMER-2017
SOLUTIONS
277528
A $25 \%$ solution of cane-sugar (mol mass $=342$ $\mathrm{g} \mathrm{mol}^{-1}$ ) is isotonic with $5 \%$ solution of a substance $A$. Then find the molecular weight of A?
1 $6.84 \mathrm{~g} \mathrm{~mol}^{-1}$
2 $68.4 \mathrm{~g} \mathrm{~mol}^{-1}$
3 $25 \mathrm{~g} \mathrm{~mol}^{-1}$
4 $684 \mathrm{~g} \mathrm{~mol}^{-1}$
Explanation:
Given, $25 \%$ solution of cane sugar means $25 \mathrm{~g}$ of solute present in $100 \mathrm{~mL}$ of solution. $\therefore$ For isotonic solutions, $\prod_{1}=\prod_{2}$ or $\quad \mathrm{C}_{1}=\mathrm{C}_{2}$ $\begin{aligned} & \frac{25}{342 \times 0.1}=\frac{5}{\mathrm{M}_{2} \times 0.1} \\ & \mathrm{M}_{2}=\frac{342}{5}=68.4 \mathrm{~g} / \mathrm{mol} \end{aligned}$
AP-EAMCET 25-08-2021 Shift - I
SOLUTIONS
277530
Which of the following pairs of solutions is expected to be isotonic at same temperature?
1 $0.1 \mathrm{M}$ urea and $0.1 \mathrm{M} \mathrm{NaCl}$
2 $0.1 \mathrm{M}$ glucose and $0.2 \mathrm{M} \mathrm{NaCl}$
3 $0.1 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}$
4 $0.1 \mathrm{M} \mathrm{NaCl}$ and $0.1 \mathrm{M} \mathrm{KNO}_{3}$
Explanation:
Two solutions of same osmotic pressure (П) are isotonic solutions. $\begin{aligned} & \prod_{1}(0.1 \mathrm{M} \mathrm{NaCl})=0.1 \times \mathrm{RT} \times 2 \\ & \prod_{2}\left(0.1 \mathrm{M} \mathrm{KNO}_{3}\right)=0.1 \times \mathrm{RT} \times 2 \end{aligned}$ So, $\Pi_{1}(0.1 \mathrm{M} \mathrm{NaCl})=\prod_{2}\left(0.1 \mathrm{M} \mathrm{KNO}_{3}\right)$ at same temperature.
