What would be the osmotic pressure at 25 c of an aqueous solution containing 1.95 g of sucrose?
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Secondary SchoolChemistry 5 points
What would be the osmotic pressure at 25°C of an aqueous solution containing 1.95g of sucrose present in 150 ml of solution?
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prafullsupe78
prafullsupe78 Expert
Hi mate.
Thanks for asking this question,
Here is your answer,
Given :
T = 25c = 298k
V = 150 ml.
n = w/m = 1.95 / 342 = 0.005 moles.
R = 8.314 J/k.mol.
\pi \: =
We know that,
\pi \: v \: = \: n \: r \: t
Therefore,
\pi \ \times 150 \: = \: 0.005 \times 8.314 \times 298
\pi \: \times 150 = \: 5 \times {10}^{? - 3 } \times 8.314 \times \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 298
\pi \: = \: 0.33 \times {10}^{? - 2} \times 2477.5
\pi \: = \: 3.3 \times {10}^{? - 3} \times 2477.5
\pi \: = 8175.98 \: \times {10}^{ - 3} pa.
T=25 C
=298 K
molar mass of sucrose=342 gm
weight of sucrose taken=1.95 gm
let V=1 L (since volume isn't given so i have taken it to be 1 L)
i=1
R=0.082
π=iCRT
=1*1.95/342*300*0.082
=0.14 atm
Hope this helps.