An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of solute?
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Given, p°A (water) = 1.013 bar
WB = 2g
p°A - pA / p°A = xB = (WB / MB) / (WA/MA) + (WB/MB)
⇒ 1.013 - 1.004 / 1.013 = (2/MB) / (98/18) [∵ WB/MB <<< WA/MA]
⇒ MB = 2×18×1.013 / 0.009×98
= 41.35 g mol^-1
Hope it helps!!!
WB = 2g
p°A - pA / p°A = xB = (WB / MB) / (WA/MA) + (WB/MB)
⇒ 1.013 - 1.004 / 1.013 = (2/MB) / (98/18) [∵ WB/MB <<< WA/MA]
⇒ MB = 2×18×1.013 / 0.009×98
= 41.35 g mol^-1
Hope it helps!!!
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