NEET Test Series from KOTA - 10 Papers In MS WORD
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SOLUTIONS
277536
Assuming the compounds to be completely dissociated in aqueous solution, identify the pair of the solution that can be expected to be isotonic at the same temperature.
1 $0.01 \mathrm{M}$ urea and $0.01 \mathrm{M} \mathrm{NaCl}$
2 $0.02 \mathrm{M} \mathrm{NaCl}$ and $0.01 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}$
3 $0.03 \mathrm{M} \mathrm{NaCl}$ and $0.02 \mathrm{M} \mathrm{MgCl}_{2}$
4 $0.01 \mathrm{M}$ sucrose and $0.02 \mathrm{M}$ glucose
Explanation:
In the isotonic solution at same temperature- $\Pi_{1}=\prod_{2}$ $\mathrm{i}_{1} \mathrm{C}_{1} \mathrm{RT}=\mathrm{i}_{2} \mathrm{C}_{2} \mathrm{RT}$ $\mathrm{i}_{1} \mathrm{C}_{1}=\mathrm{i}_{2} \mathrm{C}_{2}$ For $0.03 \mathrm{M} \mathrm{NaCl} \quad i_{1}=2$ $\mathrm{i}_{1} \mathrm{C}_{1}=2 \times 0.03$ $=0.06$ For $0.02 \mathrm{M} \mathrm{MgCl}_{2} \quad \mathrm{i}_{2}=3$ $\mathrm{i}_{2} \mathrm{C}_{2}=3 \times 0.02$ $=0.06$ Therefore, $0.03 \mathrm{M} \mathrm{NaCl}$ and $0.02 \mathrm{M} \mathrm{MgCl}_{2}$ are isotonic.
WB-JEE-2017
SOLUTIONS
277537
$5 \%$ solution of cane sugar is isotonic with $0.877 \%$ of $X$. The molecular weight of substance $X$ is
277538
$\quad 0.06 \%(\mathrm{w} / \mathrm{v})$ aqueous solution of urea is isotonic with
1 $0.06 \%$ glucose solution
2 $0.6 \%$ glucose solution
3 $0.01 \mathrm{M}$ glucose solution
4 $0.1 \mathrm{M}$ glucose solution
Explanation:
$0.06 \% \mathrm{w} / \mathrm{v}$ solution means $100 \mathrm{ml} 06$ solution contains 0.06 gm of urea- Moles of urea $=\frac{0.06}{60}=0.001 \mathrm{~mol}$ Molarity of urea solution $=\frac{0.001}{100} \times 1000$ $=0.01 \mathrm{M}$
Karnataka-CET-2015
SOLUTIONS
277539
The molar mass of a solute $X$ in $\mathrm{g} \mathrm{mol}^{-1}$, if its $1 \%$ solution is isotonic with a $5 \%$ solution of cane sugar (molar mass $\left.=342 \mathrm{~g} \mathrm{~mol}^{-1}\right)$, is
1 68.4
2 34.2
3 136.2
4 171.2
Explanation:
0.6 urea has molarity $=\frac{0.6}{60} \times \frac{1000}{100}$ $=0.1 \mathrm{M}$ glucose {|c|c|c|} | | List-I | List-II | |---|---|---| |$$ | $0.1 $ | $0.2 $ | |B | $0.1 \%( / ) $ | $10 \%( / ) $ | |C | $18 ^{-1}$ glucose | $34.2 ^{-1}$ sucrose | |$$ | $20 \%( / )$ | $10 \%( / )$ | | 705. Compartments $A$ and $B$ have the following combinations solution Indicate the number of solutions which is/are isotonic (a) I only (b) III only (c) IV only (d) II only Ans. b In the isotonic solution having the same molarity. Molarity of $18 \mathrm{~g}$ glucose $=\frac{18}{180 \times 1}$ $=\frac{18}{180}=0.1 \mathrm{~g} / \mathrm{L}$ Molarity of $34.2 \mathrm{~g}$ sucrose $=\frac{34.2}{342 \times 1}$ $=\frac{34.2}{342}=0.1 \mathrm{~g} / \mathrm{L}$ The concentration of $18 \mathrm{~g} \mathrm{~L}^{-1}$ glucose is same to $34.2 \mathrm{~g}$ $\mathrm{L}^{-1}$ sucrose.
277536
Assuming the compounds to be completely dissociated in aqueous solution, identify the pair of the solution that can be expected to be isotonic at the same temperature.
1 $0.01 \mathrm{M}$ urea and $0.01 \mathrm{M} \mathrm{NaCl}$
2 $0.02 \mathrm{M} \mathrm{NaCl}$ and $0.01 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}$
3 $0.03 \mathrm{M} \mathrm{NaCl}$ and $0.02 \mathrm{M} \mathrm{MgCl}_{2}$
4 $0.01 \mathrm{M}$ sucrose and $0.02 \mathrm{M}$ glucose
Explanation:
In the isotonic solution at same temperature- $\Pi_{1}=\prod_{2}$ $\mathrm{i}_{1} \mathrm{C}_{1} \mathrm{RT}=\mathrm{i}_{2} \mathrm{C}_{2} \mathrm{RT}$ $\mathrm{i}_{1} \mathrm{C}_{1}=\mathrm{i}_{2} \mathrm{C}_{2}$ For $0.03 \mathrm{M} \mathrm{NaCl} \quad i_{1}=2$ $\mathrm{i}_{1} \mathrm{C}_{1}=2 \times 0.03$ $=0.06$ For $0.02 \mathrm{M} \mathrm{MgCl}_{2} \quad \mathrm{i}_{2}=3$ $\mathrm{i}_{2} \mathrm{C}_{2}=3 \times 0.02$ $=0.06$ Therefore, $0.03 \mathrm{M} \mathrm{NaCl}$ and $0.02 \mathrm{M} \mathrm{MgCl}_{2}$ are isotonic.
WB-JEE-2017
SOLUTIONS
277537
$5 \%$ solution of cane sugar is isotonic with $0.877 \%$ of $X$. The molecular weight of substance $X$ is
277538
$\quad 0.06 \%(\mathrm{w} / \mathrm{v})$ aqueous solution of urea is isotonic with
1 $0.06 \%$ glucose solution
2 $0.6 \%$ glucose solution
3 $0.01 \mathrm{M}$ glucose solution
4 $0.1 \mathrm{M}$ glucose solution
Explanation:
$0.06 \% \mathrm{w} / \mathrm{v}$ solution means $100 \mathrm{ml} 06$ solution contains 0.06 gm of urea- Moles of urea $=\frac{0.06}{60}=0.001 \mathrm{~mol}$ Molarity of urea solution $=\frac{0.001}{100} \times 1000$ $=0.01 \mathrm{M}$
Karnataka-CET-2015
SOLUTIONS
277539
The molar mass of a solute $X$ in $\mathrm{g} \mathrm{mol}^{-1}$, if its $1 \%$ solution is isotonic with a $5 \%$ solution of cane sugar (molar mass $\left.=342 \mathrm{~g} \mathrm{~mol}^{-1}\right)$, is
1 68.4
2 34.2
3 136.2
4 171.2
Explanation:
0.6 urea has molarity $=\frac{0.6}{60} \times \frac{1000}{100}$ $=0.1 \mathrm{M}$ glucose {|c|c|c|} | | List-I | List-II | |---|---|---| |$$ | $0.1 $ | $0.2 $ | |B | $0.1 \%( / ) $ | $10 \%( / ) $ | |C | $18 ^{-1}$ glucose | $34.2 ^{-1}$ sucrose | |$$ | $20 \%( / )$ | $10 \%( / )$ | | 705. Compartments $A$ and $B$ have the following combinations solution Indicate the number of solutions which is/are isotonic (a) I only (b) III only (c) IV only (d) II only Ans. b In the isotonic solution having the same molarity. Molarity of $18 \mathrm{~g}$ glucose $=\frac{18}{180 \times 1}$ $=\frac{18}{180}=0.1 \mathrm{~g} / \mathrm{L}$ Molarity of $34.2 \mathrm{~g}$ sucrose $=\frac{34.2}{342 \times 1}$ $=\frac{34.2}{342}=0.1 \mathrm{~g} / \mathrm{L}$ The concentration of $18 \mathrm{~g} \mathrm{~L}^{-1}$ glucose is same to $34.2 \mathrm{~g}$ $\mathrm{L}^{-1}$ sucrose.
277536
Assuming the compounds to be completely dissociated in aqueous solution, identify the pair of the solution that can be expected to be isotonic at the same temperature.
1 $0.01 \mathrm{M}$ urea and $0.01 \mathrm{M} \mathrm{NaCl}$
2 $0.02 \mathrm{M} \mathrm{NaCl}$ and $0.01 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}$
3 $0.03 \mathrm{M} \mathrm{NaCl}$ and $0.02 \mathrm{M} \mathrm{MgCl}_{2}$
4 $0.01 \mathrm{M}$ sucrose and $0.02 \mathrm{M}$ glucose
Explanation:
In the isotonic solution at same temperature- $\Pi_{1}=\prod_{2}$ $\mathrm{i}_{1} \mathrm{C}_{1} \mathrm{RT}=\mathrm{i}_{2} \mathrm{C}_{2} \mathrm{RT}$ $\mathrm{i}_{1} \mathrm{C}_{1}=\mathrm{i}_{2} \mathrm{C}_{2}$ For $0.03 \mathrm{M} \mathrm{NaCl} \quad i_{1}=2$ $\mathrm{i}_{1} \mathrm{C}_{1}=2 \times 0.03$ $=0.06$ For $0.02 \mathrm{M} \mathrm{MgCl}_{2} \quad \mathrm{i}_{2}=3$ $\mathrm{i}_{2} \mathrm{C}_{2}=3 \times 0.02$ $=0.06$ Therefore, $0.03 \mathrm{M} \mathrm{NaCl}$ and $0.02 \mathrm{M} \mathrm{MgCl}_{2}$ are isotonic.
WB-JEE-2017
SOLUTIONS
277537
$5 \%$ solution of cane sugar is isotonic with $0.877 \%$ of $X$. The molecular weight of substance $X$ is
277538
$\quad 0.06 \%(\mathrm{w} / \mathrm{v})$ aqueous solution of urea is isotonic with
1 $0.06 \%$ glucose solution
2 $0.6 \%$ glucose solution
3 $0.01 \mathrm{M}$ glucose solution
4 $0.1 \mathrm{M}$ glucose solution
Explanation:
$0.06 \% \mathrm{w} / \mathrm{v}$ solution means $100 \mathrm{ml} 06$ solution contains 0.06 gm of urea- Moles of urea $=\frac{0.06}{60}=0.001 \mathrm{~mol}$ Molarity of urea solution $=\frac{0.001}{100} \times 1000$ $=0.01 \mathrm{M}$
Karnataka-CET-2015
SOLUTIONS
277539
The molar mass of a solute $X$ in $\mathrm{g} \mathrm{mol}^{-1}$, if its $1 \%$ solution is isotonic with a $5 \%$ solution of cane sugar (molar mass $\left.=342 \mathrm{~g} \mathrm{~mol}^{-1}\right)$, is
1 68.4
2 34.2
3 136.2
4 171.2
Explanation:
0.6 urea has molarity $=\frac{0.6}{60} \times \frac{1000}{100}$ $=0.1 \mathrm{M}$ glucose {|c|c|c|} | | List-I | List-II | |---|---|---| |$$ | $0.1 $ | $0.2 $ | |B | $0.1 \%( / ) $ | $10 \%( / ) $ | |C | $18 ^{-1}$ glucose | $34.2 ^{-1}$ sucrose | |$$ | $20 \%( / )$ | $10 \%( / )$ | | 705. Compartments $A$ and $B$ have the following combinations solution Indicate the number of solutions which is/are isotonic (a) I only (b) III only (c) IV only (d) II only Ans. b In the isotonic solution having the same molarity. Molarity of $18 \mathrm{~g}$ glucose $=\frac{18}{180 \times 1}$ $=\frac{18}{180}=0.1 \mathrm{~g} / \mathrm{L}$ Molarity of $34.2 \mathrm{~g}$ sucrose $=\frac{34.2}{342 \times 1}$ $=\frac{34.2}{342}=0.1 \mathrm{~g} / \mathrm{L}$ The concentration of $18 \mathrm{~g} \mathrm{~L}^{-1}$ glucose is same to $34.2 \mathrm{~g}$ $\mathrm{L}^{-1}$ sucrose.
277536
Assuming the compounds to be completely dissociated in aqueous solution, identify the pair of the solution that can be expected to be isotonic at the same temperature.
1 $0.01 \mathrm{M}$ urea and $0.01 \mathrm{M} \mathrm{NaCl}$
2 $0.02 \mathrm{M} \mathrm{NaCl}$ and $0.01 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}$
3 $0.03 \mathrm{M} \mathrm{NaCl}$ and $0.02 \mathrm{M} \mathrm{MgCl}_{2}$
4 $0.01 \mathrm{M}$ sucrose and $0.02 \mathrm{M}$ glucose
Explanation:
In the isotonic solution at same temperature- $\Pi_{1}=\prod_{2}$ $\mathrm{i}_{1} \mathrm{C}_{1} \mathrm{RT}=\mathrm{i}_{2} \mathrm{C}_{2} \mathrm{RT}$ $\mathrm{i}_{1} \mathrm{C}_{1}=\mathrm{i}_{2} \mathrm{C}_{2}$ For $0.03 \mathrm{M} \mathrm{NaCl} \quad i_{1}=2$ $\mathrm{i}_{1} \mathrm{C}_{1}=2 \times 0.03$ $=0.06$ For $0.02 \mathrm{M} \mathrm{MgCl}_{2} \quad \mathrm{i}_{2}=3$ $\mathrm{i}_{2} \mathrm{C}_{2}=3 \times 0.02$ $=0.06$ Therefore, $0.03 \mathrm{M} \mathrm{NaCl}$ and $0.02 \mathrm{M} \mathrm{MgCl}_{2}$ are isotonic.
WB-JEE-2017
SOLUTIONS
277537
$5 \%$ solution of cane sugar is isotonic with $0.877 \%$ of $X$. The molecular weight of substance $X$ is
277538
$\quad 0.06 \%(\mathrm{w} / \mathrm{v})$ aqueous solution of urea is isotonic with
1 $0.06 \%$ glucose solution
2 $0.6 \%$ glucose solution
3 $0.01 \mathrm{M}$ glucose solution
4 $0.1 \mathrm{M}$ glucose solution
Explanation:
$0.06 \% \mathrm{w} / \mathrm{v}$ solution means $100 \mathrm{ml} 06$ solution contains 0.06 gm of urea- Moles of urea $=\frac{0.06}{60}=0.001 \mathrm{~mol}$ Molarity of urea solution $=\frac{0.001}{100} \times 1000$ $=0.01 \mathrm{M}$
Karnataka-CET-2015
SOLUTIONS
277539
The molar mass of a solute $X$ in $\mathrm{g} \mathrm{mol}^{-1}$, if its $1 \%$ solution is isotonic with a $5 \%$ solution of cane sugar (molar mass $\left.=342 \mathrm{~g} \mathrm{~mol}^{-1}\right)$, is
1 68.4
2 34.2
3 136.2
4 171.2
Explanation:
0.6 urea has molarity $=\frac{0.6}{60} \times \frac{1000}{100}$ $=0.1 \mathrm{M}$ glucose {|c|c|c|} | | List-I | List-II | |---|---|---| |$$ | $0.1 $ | $0.2 $ | |B | $0.1 \%( / ) $ | $10 \%( / ) $ | |C | $18 ^{-1}$ glucose | $34.2 ^{-1}$ sucrose | |$$ | $20 \%( / )$ | $10 \%( / )$ | | 705. Compartments $A$ and $B$ have the following combinations solution Indicate the number of solutions which is/are isotonic (a) I only (b) III only (c) IV only (d) II only Ans. b In the isotonic solution having the same molarity. Molarity of $18 \mathrm{~g}$ glucose $=\frac{18}{180 \times 1}$ $=\frac{18}{180}=0.1 \mathrm{~g} / \mathrm{L}$ Molarity of $34.2 \mathrm{~g}$ sucrose $=\frac{34.2}{342 \times 1}$ $=\frac{34.2}{342}=0.1 \mathrm{~g} / \mathrm{L}$ The concentration of $18 \mathrm{~g} \mathrm{~L}^{-1}$ glucose is same to $34.2 \mathrm{~g}$ $\mathrm{L}^{-1}$ sucrose.
