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Question

On a certain hill station pure water is found to boil at 950C. How many grams of NaClis required so that water boils at 1000C

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Solution

Boiling point of pure water(Tob) = 95oCBoiling point of the solution (Tb) =100oCElevation in boiling point (ΔTb)= Tb-Tob = 100-95=5oCΔTb = i Kbmi=Van't Hoff factor = for strong electrolyte it is equal to the number of ions it breaks into i.e. for NaCl =2m=molality = moles of solute in 1000 g of solventKb=molal elevation constant = 0.512 oC/molalLet mass of water = 100 gHence, 5 =2×0.512×WM×1000mass of solvent (g)W = mass of NaClM = Molar mass of NaCl = 23 + 35.5 = 58.5 g5 = 2×0.512×W58.5×10001000W = 5×58.5×10000.512×1000×2 = 285.6g

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