How many grams of must be added to 250 of water to lower the vapor pressure by 1

Question

How many grams of must be added to 250 of water to lower the vapor pressure by 1.30 at 40 assuming complete dissociation? The vapor pressure of water at 40 is 55.3 .

Explanation / Answer

according to raoults law ... Po -Ps/Po = X2 but since NaBr is also dissociating ...so we have to introduce a vant hoff factor , i hence raoluts law modifies to Po -Ps/Po = i X2 NaBr -----> Na+ + Br- so i =2 so ..Po -Ps/Po = 2X2 where Po = vapour pressure of pure solvent = 55.3 mm Hg Ps = vapor pressure of solution = 55.3-1.3 = 54 mm Hg X2 = mole fraction of solute which is NaBr ... let amount of NaBr taken be w and molecular mass of NaBr = 103 g/mole then no. of mole of NaBr = w/103 amount of water = 250 g and molecular mass of water = 18 g/mole so no. of moles of water = 250/18 = 13.89 total no. of moles = 13.89 + w/103 so mole fraction of NaBr = no. of moles of NaBr/total no. of moles X2 = w/103/(w/103 + 13.89) so Po-Ps/Po = 2X2 putting the repective values ..and solving for w 55.3-54/55.3 = 2 X w/103/(w/103 + 13.89) 1.3/55.3 = 2 X w/103 / (w+1430.67/103) 0.024 = 2w/w+1430.67 0.024(w+1430.67) = 2w 0.024w + 34.34 = 2w 2w-0.024w = 34.34 1.976w = 34.34 w = 17.38

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