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A solution of 1.25 g of a certain non-electrolyte in 20.0 g of water freezes at 271.94 K. Calculate the molecular mass of the solute, `K_(f)` for water is 1.86 K/m. |
Answer» Correct Answer - 109.67 g `mol^(-1)` `W_(B)=1.25g, W_(A)=0.02 kg, K_(f)-=1.86" K kg mol"^(-1),T_(f)^(@)=273 K` `T_(f)=271.94 K, DeltaT_(f)=273-271.94=1.06 K, M_(B)=?` `M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))=((1.86" K kg mol"^(-1))xx(1.25 g))/((1.06 K)xx(0.02 kg))=109.67" g mol"^(-1).` |
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