The relationship between the value of osmotic pressure - of the solution obtained by dissolving 6g L -1 of acetic acid - 1 and 7-45g L -1 of KCl - 2 is-

The correct option is A π_ 1 π=l_1C_1RT =660×(1+α)RT #160; #160; #160; #160; (α lt; 1) ⇒1+ α10RT =π_2=l_2C_2RT =7.457.45 ...

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Standard XII

Chemistry

Relative Lowering of Vapour Pressure

The relations...

Question

The relationship between the value of osmotic pressure $(π)$ of the solution obtained by dissolving 6g $L^{- 1}$ of acetic acid $(π_{1})$ and 7.45g $L^{- 1}$ of $K C l$ $(π_{2})$ is:

π_{1} < π_{2}

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π_{1} > π_{2}

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π_{1} = π_{2}

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\frac{π_{1}}{π_{1} + π_{2}} = \frac{π_{2}}{π_{1} + π_{2}}

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Solution

The correct option is A $π_{1} < π_{2}$
$π = l_{1} C_{1} R T$
$= \frac{6}{60} \times (1 + α) R T$ $(α < 1)$
$\Rightarrow \frac{1 + α}{10} R T$
$= π_{2} = l_{2} C_{2} R T$
$= \frac{7.45}{7.45} \times 2 \times R T$
$= \frac{1}{5} R T$
$∴ π_{2} > π_{2}$

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The relationship between the value of osmotic pressure (π) of the solution obtained by dissolving 6g L−1 of acetic acid (π1) and 7.45g L−1 of KCl (π2) is:

The correct option is A π1<π2π=l1C1RT=660×(1+α)RT (α<1)⇒1+α10RT=π2=l2C2RT=7.457.45×2×RT=15RT∴π2>π2

The relationship between the value of osmotic pressure $(π)$ of the solution obtained by dissolving 6g $L^{- 1}$ of acetic acid $(π_{1})$ and 7.45g $L^{- 1}$ of $K C l$ $(π_{2})$ is:

The correct option is A $π_{1} < π_{2}$
$π = l_{1} C_{1} R T$
$= \frac{6}{60} \times (1 + α) R T$ $(α < 1)$
$\Rightarrow \frac{1 + α}{10} R T$
$= π_{2} = l_{2} C_{2} R T$
$= \frac{7.45}{7.45} \times 2 \times R T$
$= \frac{1}{5} R T$
$∴ π_{2} > π_{2}$