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New answer posted
5 months agoContributor-Level 9
At a fixed temperature, having more vapour pressure as compared to . So, intermolecular interaction is lower as compared to .
New answer posted
5 months agoContributor-Level 10
Vapour pressure over solvent is greater than that over solution.
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5 months agoContributor-Level 10
The above mixture will show positive deviation from Raoult's law.
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5 months agoContributor-Level 9
P (CO? ) = K? X (CO? )
X (CO? ) = P (CO? )/K? = 0.835 / (1.67 * 10³) = 0.5 * 10? ³
X (CO? ) = n (CO? )/ (n (CO? ) + n (H? O) ≈ n (CO? )/n (H? O) (since n (CO? ) << n (H? O)
n (H? O) in 0.9L = 900g/18gmol? ¹ = 50 mol
n (CO? ) = X (CO? ) * n (H? O) = 0.5 * 10? ³ * 50 = 25 * 10? ³ moles = 25 mmol
New answer posted
5 months agoContributor-Level 10
? = CRT
2.42 * 10? ³ = ( (1.46/M_polymer) / 0.1 ) * 0.083 * 300
M_polymer = (1.46 * 0.083 * 300) / (2.42 * 10? ³ * 0.1)
= 14.96 * 10? gm = 15 * 10? gm
New answer posted
5 months agoContributor-Level 10
ΔTb = 0.6 K
ΔTb = Kb * m = Kb * (wt. * 1000)/ (mol wt. * Wsolvent (gm)
0.6 = 5 * (3 * 1000)/ (mol wt. * 100)
∴ mol wt. = (15 * 10)/ (0.6) = 1500/6
= 250 g/mole
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