The vapour pressure of water at 293 k is 2438 Pa and the vapour pressure of an aqueous solution is 2395.8 Pa. If solution density is 1080 $k g / m^{3}$ at 313 K. Calculate the osmotic pressure at 313 K. The molecular weight of solute =60.

24.54

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24.54 \times 10^{5}

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0.955

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None of these

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Solution

The correct option is C $24.54 \times 10^{5}$ Pa
$\frac{P^{o} - P_{s}}{P_{s}} = \frac{n \times 18}{W}$

$∴ \frac{n}{W} = \frac{(2438 - 2395.8)}{2438 \times 18} = 9.616 \times 10^{- 4}$

Concentration, $C = \frac{n}{W} \times d = 9.616 \times 10^{- 4} \times 1080 = 1.038$

Osmotic Pressure $= C R T$

$⟹$ Osmotic Pressure $= 1.038 \times 0.0821 \times 313$

$⟹$ Osmotic Pressure $= 24.54$ $a t m$

$⟹$ Osmotic Pressur $= 24.54 \times 10^{5}$ $P_{a}$

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The vapour pressure of water at 293 k is 2438 Pa and the vapour pressure of an aqueous solution is 2395.8 Pa. If solution density is 1080 kg/m3 at 313 K. Calculate the osmotic pressure at 313 K. The molecular weight of solute =60.

The correct option is C 24.54×105PaPo−PsPs=n×18W ∴nW=(2438−2395.8)2438×18=9.616×10−4 Concentration, C=nW×d=9.616×10−4×1080=1.038 Osmotic Pressure =CRT ⟹ Osmotic Pressure =1.038×0.0821×313 ⟹ Osmotic Pressure =24.54 atm ⟹ Osmotic Pressur =24.54×105 Pa

The vapour pressure of water at 293 k is 2438 Pa and the vapour pressure of an aqueous solution is 2395.8 Pa. If solution density is 1080 $k g / m^{3}$ at 313 K. Calculate the osmotic pressure at 313 K. The molecular weight of solute =60.