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Some chemistry problems

  1. Feb 3, 2005 #1
    Problem 1:

    How many grams of a solution that is 12.5% by mass [tex]AgNO_3[/tex] would contain 0.400 mol of [tex]AgNO_3[/tex]?

    The back of the book says that the answer is 544 g solution, and I got the answer, but only by randomly multiplying and subtracting the percentage, number of moles, molar mass, and the number 2. All I'd really like for this one is the equation to solve it. The book instructed mass% = [tex]\frac{g solute}{g solution} \times 100[/tex] as the main equation to use for problems like this. Is this right, because I tried converting the equation, but I got a different answer.

    Problem 2:

    What is the molar mass of a compound if 4.80 g of the compound dissolved in 22.0 g of [tex]H_2O[/tex] gives a solution that freezes at [tex]-2.50^\circ C[/tex]?


    Thanks. :)
     
  2. jcsd
  3. Feb 3, 2005 #2

    dextercioby

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    The first problem is elementary.Compute the molar mass of silver nitrate and then knowing the # of moles & the molar mass,u can find the mass.
    Then simply use the definition of massic concentration of a sollution...

    Daniel.
     
  4. Feb 4, 2005 #3

    Gokul43201

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    As for Q2, surely you must have completed Colligative Properties, for this to be in your homework. On the off chance that you haven't, look up this chapter in your text and find the formula for the depression of freezing point. You will also need the definition of 'molality' but this will be found in the same chapter or in the "Mole Concept" (?) chapter.

    Finally, there's always Google.
     
  5. Feb 4, 2005 #4

    cepheid

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    Problem 1: What's this about mindlessly applying equations? The problem is easy to understand. You know this: 0.400 moles of silver nitrate make up 12.5% (1/8) of the mass of the solution. Just calculate the mass of 0.400 mol of silver nitrate. The mass of the whole solution is 8 times that.

    Problem 2: Please post your attempt so far.
     
  6. Feb 5, 2005 #5

    saltydog

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    Problem 2:

    Well, if the freezing point depression constant is 1.86, that means for every mole of dissolved particles in a kg of water, the freezing pt. will drop by 1.86 degrees. If the solute is ionic, you need to consider all the ions (like NaCl would be twice as many, magnesium chloride, 3 right?). Assume the solute is not ionic: Since the problem says 2.5, that means the solution must be 2.5/1.86 or 1.34 m right?

    Now, the problem said that 4.8g were dissolved in 22 grams (0.022kg) of water, then the following must be true:

    [tex] 1.34=\frac{\frac{22}{x}}{0.022} [/tex]

    where x is the mol.wgt.

    That is, 1.34 is equal to the number of moles (22/x) per kg (0.022) of water. You can finish it.

    Salty
     
  7. Feb 6, 2005 #6
    wow, thanks everyone. i was seeing it much harder than it really was. :blushing:
     
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