Using data from the table below, calculate the freezing and boiling points of 2.

Question

Using data from the table below, calculate the freezing and boiling points of 2.04 g KBr and 4.82 g glucose (C6 H12O6) in 188 g H2O. Given: Kb of H2O: 0.51 C/m; Kf of H2O: 1.86 C/m. I'm not really looking for the answer (though I won't complain if you provide it), I'm just confused as to what I need to do if I have a solution with two solutes. I know I need to use the equation Delta(Tb)=Kb*m*i and Delta(Tf)=Kf*m*i. I know how to figure it out if it's only 1 solute but I'm confused as to what to do if I have two solutes. (By the way, I put this under my course ENGR 202; it's actually for ENGR 106 but it wasn't wanting to load that course. Anyways, that's trivial.

Explanation / Answer

molality of KBr = (2.04 / 119) / 0.188 = 0.091 m......molality of glucose = 4.82 / 180 = 0.0268 m......i for KBr = 2(K+ + Br-)......i for Glucose = 1 ....... Delta T(b) = 0.51 x 1 x 0.0268 + 0.51 x 2 x 0.091 = 0.1065 C ..............Delta T(f) = 1.86 x 1 x 0.0268 + 1.86 x 2 x 0.091 = 0.388 C. The molality of the two solutes need to be added in case of two solutes

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