<p>Kudos to anyone who can make any headway on this problem!!!</p>
<p>A 2.60g sample of an unknown metallic halide(fmu=130g/mol) dissociates completely in 150g of water. If the boiling point of the solution is found to be 100.205C, what is the charge of the metal ion? (The molal boiling point elevation constant for water is 0.512C/molal.)</p>
<p>what in the world is fmu?</p>
<p>another word for molar mass, basically.</p>
<p>Basically, you are using the equation delta(T)=i<em>m</em>k.
delta(t)=change in boiling point=100.205-100=0.205<em>C
i=van hoff factor
m=molality, can be calculated
k=boiling point elevation constant for solvent=0.512</em>C/mol</p>
<p>For molality, m=(mol unknown)/(kg water)
mol unknown=2.6g/(130g/mol)=0.02
molality=m=(0.02<em>mol)/(.150</em>kg)=0.1333*mol/kg</p>
<p>From boiling pt ele. eqn:
0.205<em>C=i</em>0.1333<em>mol/kg</em>0.512*C/molal
i=3</p>
<p>The van hoff factor is the number of parts that the molecule breaks into.</p>
<p>It is given that the unknown is a metallic halide, so it is in the form of A B_x, where A is a metal, B is a halogen, and _x denotes a subscript on B.</p>
<p>So, the molecule will break into x+1 parts. x+1=3, x=2.</p>
<p>Thus, the charge on the metal is +2</p>
<p>ahhhhh! my head is splitting. i was once proud that im good in chem. but this is a killer.</p>
<p>what do i care... i got a 5... gl though</p>
<p>"Basically, you are using the equation delta(T)=i<em>m</em>k.
delta(t)=change in boiling point=100.205-100=0.205<em>C
i=van hoff factor
m=molality, can be calculated
k=boiling point elevation constant for solvent=0.512</em>C/mol</p>
<p>For molality, m=(mol unknown)/(kg water)
mol unknown=2.6g/(130g/mol)=0.02
molality=m=(0.02<em>mol)/(.150</em>kg)=0.1333*mol/kg</p>
<p>From boiling pt ele. eqn:
0.205<em>C=i</em>0.1333<em>mol/kg</em>0.512*C/molal
i=3</p>
<p>The van hoff factor is the number of parts that the molecule breaks into.</p>
<p>It is given that the unknown is a metallic halide, so it is in the form of A B_x, where A is a metal, B is a halogen, and _x denotes a subscript on B.</p>
<p>So, the molecule will break into x+1 parts. x+1=3, x=2.</p>
<p>Thus, the charge on the metal is +2."</p>
<p>Uhh, who taught you this formula??? I am so screwed, I have no idea what the van hoff factor is.</p>
<p>I am gonna invade this thread. Any recs on what to use to study and get a 4, or 5 on the test? My teacher doesn't know how to teach, so I am cracking the review books early, I'll memorize every single one of them if I have to. Thanks a lot guys.</p>
<p>man<em>on</em>fire, it is actually called the "van't Hoff factor," my mistake. The formula is encountered when studying colligative properties.</p>
<p>Catch-22, I used PR and it helped. However, I ended up with a 2 (didn't know much about chemistry).</p>
<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>
<p>To solve this problem, I need to give a more elaborate definition of the van't Hoff factor (labeled by i). The van't Hoff factor is the ratio of solute particles in the solution to the number of solute particles put in. In examples involving strong electrolytes (which dissolve completely in water), this comes out to a whole number because there is a whole number of particles in soln for each initial solute particle ionized. However, in the case of a weak electrolyte, the solute only partially dissolves and many ionic bonds are left intact.</p>
<p>In your example, we are using a weak electrolyte, HY: this will not dissolve completely in water. Therefore, i will be between 1 and 2. Due to the nature of weak electrolytes, you can expect i to be closer to 1.</p>
<p>Use the formula, delta(t)=i<em>m</em>k to find i=1.15</p>
<p>Now look at the definition of the van't Hoff factor. i represents a yield essentially. In this case for every HF particle put in, we get 1.15 particles out (on average). That means some of the ionic bonds were broken. If we are getting 15% extra particles, 15% of the ionic bonds were broken. Since ionization is the process of breaking ionic bonds, you have your answer.</p>
<p>Hope this helps.</p>
<p>thanks, averagemathgeek!</p>
<p>r u taking chem 2 this yr? it seems like u know the entire chem textbook!</p>
<p>I am taking Chemistry at a university (I'm a college student). I know very little chemistry off the top of my head: I just read my notes or the book when I need something.</p>
<p>Here's some other tough cookies to crack!!!</p>
<p>If vinegar is 5.0% acetic acid(HC2H3O2, mol. wt=60g/mol) in water, what is the molality of vinegar?</p>
<p>An aqueous solution of sucrose(C12H12O11, mol. wt=342g/mol) is 20% sucrose by mass. Calculate the vapor pressure of the solution at 100C.</p>