SAT II Chemistry

Test Preparation
1

<p>ohhhhhhh, can anybody help me on this question:</p>

<p>A 0.100 molality aqueous solution of a weak electroyte, HY, freezes at -0.214 degree celcius. What is the approximate percent ionization of HY? The freezing point depression constant for water is 1.86 C/m.</p>

<p>Correct anwer: 15.0%</p>

<p>i have no idea how they got this because the solution mess up.</p>

<p>plzzzzzzzzzz, anybody please help me!</p>

2

<p>use the formula: ∆T = i<em>k</em>m
i = ?
m= 0.100
k = 1.86
∆T = 0.214</p>

<p>Solving this you will get: i = 1.150. i = 2 if HY ionized completely, which means that the percent ionization is 15.0%.</p>

3

<p>Is this covered on the SAT II? I haven't seen this in the barron's or sparknotes books...</p>