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Calculate the mass of NaCI (molar mass =58.5 g `mol^(-1)`) to be dissoved in 37.2 g of water to lower the freezing point by `2^(@)C` assuming that NaCI underfoes complete dissocilation in aqueous solution (`K_(f)` for water = 1.86 K kg `mol^(-1)`). |
Answer» Correct Answer - 1.16 g `i=(DeltaT_(f))/(K_(f)xxm)=(DeltaT_(f)xxM_(B)xxM_(A))/(K_(f)xxW_(B))or W_(B)=(DeltaT_(f)xxM_(B)xxM_(A))/(ixxK_(f))` `i=2("for complete dissciation of NaCI"),K_(f)=1.86" K kg mol"^(-1)` `DeltaT_(f)=2K, M_(B)=58.5" g mol"^(-1), M_(A)=37.2 g=0.037 kg` `W_(B)=((2K)xx(58.5"gmol"^(-1))xx(0.037 kg))/(2xx(1.86" K kg mol"^(-1)))=1.16 g` |
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