You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Osmotic Pressure

The vapour pr...

Question

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

24.54 \times 10^{5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.955

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Suggest Corrections

Similar questions

Q. The vapour pressure of water at 293 K is 2338 Pa and the vapour pressure of an aqueous solution is 2295.8 Pa. The molar mass of solute is

60 g m o l^{- 1}

.The osmotic pressure at 313 K, if the solution density at this temperature is 1010

K g / m^{3}

Q. The vapour pressure of water at

293 K

2338 P a

and the vapour pressure of an aqueous solution is

2295.8 P a

. If solution density is

1010 k g / m^{3}

313 K

, calculate the osmotic pressure at

313 K

. Molecular weight of solute

= 60

Q. Benzene and toluene from two ideal solution A and B at 313 K. Solution A (total pressure

P_{A}

) contains equal mole of toluene and benzene. Solution B contains equal masses of both A (total pressure

P_{B}

). The vapour pressure of benzene and toluene are 160 and 60 mm Hg respectively at 313 K. Calculate the value of

P_{A} + P_{B}

(Round-off the answer).

QUESTION 2.19

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 k Pa at 298 K. Calculate (i) molar mass of the solute (ii) vapour pressure of water at 298 K.

Benzene and toluene form nearly ideal solution .At 313K the vapour pressure of pure benzene is 150mm Hg and of pure toluene is 50 mm Hg .calculate vapour pressure of mixture of these two containing their equal masses at 313 K .

Join BYJU'S Learning Program

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.