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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C. - Chemistry

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प्रश्न

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

संख्यात्मक
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उत्तर

Given,

Number of moles of solute dissolved (n) = `(1.0  "g")/(185,000  "g mol"^(-1)) = 1/(185,000)  "mol"`

V = 450 mL = 0.45 L,

T = 37°C = (37 + 273) K = 310 K

R = 8.314 k Pa L K−1 mol−1 = 8.314 × 103 Pa L K−1 mol−1

Osmotic pressure, `pi = "n"/"V""RT"`

= `1/(185,000)  "mol" xx 1/(0.45  "L")xx 8.314 xx 10^(3)  "Pa L K"^(-1)  "mol"^(-1)xx 310  "K"`

= 30.96 Pa

= 31 Pa (approximately)

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पाठ 1: Solutions - Intext Questions [पृष्ठ २३]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
पाठ 1 Solutions
Intext Questions | Q 1.12 | पृष्ठ २३

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