- Beaker ge test tube rack CI) Clamp C1) - Crucible Tong ( Buret Clamp (1) - 9 p

Question

- Beaker ge test tube rack CI) Clamp C1) - Crucible Tong ( Buret Clamp (1) - 9 points: . The freezing point of a solution of 1.65 g of a nonelectrolyte un known pure benzene is 6.5 oc, and the freezing-point °C/molal ed in a 20.0 g of benzene is 1.1 °C. The freezing point of constant for benzene is 5.12 a. What is the molalit y of the unknown compound in this solution? (3 pts) b. What is the molar mass of the unknown compound? (3 pts) c, suppose that, in the experiment described above, the freezing point the solution had been inaccurately recorded as 2.0 °C, instead of oc. Would the new calculated molar mass of the unknown have higher, lower, or unaffected? Briefly justiífy your response. (3 pts) 1.1 been

Explanation / Answer

Freezing point of solution = Ts = 1.1C

Mass of solute = 1.65 grams

Mass of solvent (benzene) = 20.0grams

Freezing point of Pure solvent (benzene) = T0 = 6.5C

Freezing point constant= Kf = 5.12

deltaTf= Kf x molality

T0 - Ts= kx molality

6.5-1.1 = 5.12x molality

molality = 5.4/5.12 = 1.05m

molality of unknown compound = 1.05m

b) molality = w/Mw x1000/W

where w= mass of solute, MW = molar mass of unknown compound, W= mass of solvent

1.05 = 1.65/MW x1000/20

MW= 1.65x1000/1.05x20 = 78.57 grams.

Molar mass of unknown compound = 78.57 grams.

c) Freezing point of the solution is increses, Depression in freezing point is decreases. so molalit y value is also decreases.

Molality decreases, the moleculare mass of unknouwn compound is increases.

So freezing point of solution is taken as 2.0C , molecular mass is higher value.

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