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# scan0003 - 1 A Molar mass from freezing point depression a...

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Unformatted text preview: 1% A. Molar mass from freezing point depression a. Using a piece of paper as a straightedge, draw the lines that are used to determine the freezing point of this solution. Answer is like Fig. 12A.3, p.215. b. The freezing point depression is 2 . 65°C and depression constant It, for this solvent is 20.5“C/mola1. What is the molality of the solution studied? A1: = K, c,l C/ m AT 2.65 °C (3‘ll == --— == R; 20.5 °C / molal 0m =: —— ==(0.129268m01al Cm == T 2 0.129mola f \— o\ c. The solute was dissolved in 25.58 g of so vent. How many moles of solute are in the solution? \ n \ ! wtsolvent \ M 3 5X19 ) ( 25.58 6%; ) —-——._~____ 1000 Gram / Kilogram Cn == n == c. tholveut == (0.129 Kilogram n 2: t:In tholvent = 0. 00329982 Mole n =: c... wt501vent == 0.00330 Mole d. If the mass of the solute is 0.3506 g, what i the molar mass of the solute? m_ we 0.35066ram / .-\ " moles " 0.00330m12é 93W“) MW— wt ‘_ 106.242 Gram " moles " Mole -- wt __ 106 Gram moles Mole ...
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