Boiling point elevation
When g of a non-volatile and nonelectrolyte solute was dissolved into g of pure water, the boiling point of the solution was C. Then what is the molecular weight of the solute? (The ebullioscopic constant of water is = C/m.)
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1 solution
Since we know that water boils at 100 degree celcius, adding the non-electrolye solute changes the solution's boiling from 100 to 100.26 or a net change of 0.26 degree celcius.
Change in temp for solute= molality x ebullioscope of water 0.26= molality x 0.52 Therefore, molality for this solute is 0.26/0.52=0.5 mol/kg solute
So, in 200 grams of water, the number of moles of water/kg is = 0.5 x 200/1000 = 0.1 moles.
If 6 grams of solute was dissolved in 0.1 moles of water, then calculate the number of moles of solute(molar mass) per mol of water. So we get 6/0.1=60 g