Chem 1120 2017 Laboratory Exercise: Colligati ve Properties- Molar Mass by Freez

Question

Chem 1120 2017 Laboratory Exercise: Colligati ve Properties- Molar Mass by Freezing Point Depression (p 4) Problems 1. A 0.0500 m aqueous solution of KalFe(CN)Je] has a freezing point of o.2800 C. Kfor wate 1.86 Cl m. calculate the total concentration (m) of solute particles in this solution. Note that it is not simply the m of Ks[Fe(CN)o) 2. The freezing point depression of 1.00 % by mass aqueous solution of acetic acid (CHCOOH) is 0.310 K, for water = 1.860cm. Calculate the formula mass of acetic acid based on these data. The freezing point depression of a 1.00 % by mass solution of acetic acid (CH3COOH) in benzene is 0.441° K, for benzene-5.12 °C/m; Calculate the formula mass of acetic using these data. 3. Propose an explanation why these acetic acid calculated molar masses are different

Explanation / Answer

Ans 1:

the freezing point depression = kf.m

0.2800=1.86 × m

m = 0.15

So the total concentration of solute particles is 0.15 m

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