Results 1 to 7 of 7

Math Help - Chemistry Stuff (Solutions)

  1. #1
    Newbie
    Joined
    Jul 2007
    From
    Perth, Western Australia
    Posts
    4

    Chemistry Stuff (Solutions)

    1. Calculate the number of moles of nitrate ions in 2.20L of 2.02x10-3 mol L-1 lead(II) nitrate solution.

    2. Calculate the mass of solute which must be used in order to prepare the following solution 630 mL of 1.26 molL-1 KCl (aq) from KCl (s)

    3. What volume of water must be added to 150.0mL of 1.10 molL-1 sulfuric acid solution to prepare a 0.210 mol L-1 solution?

    4. Commercial sulphuric acid contains 98.0% H2SO4 by mass and has a density of 1.84g mL-1. Calculate the concentration in mol L-1 of the solution.

    5. A 14.2 mol L-1 solution of NaOH has a density of 1.42g mL-1. What is the percentage by mass of NaOH in the solution?

    6. 150mL of a 1.01 Mol L-1 KCl solution had 11.7g of NaCl dissolved in it. Calculate the new concentration in mol L-1 of the chloride ion. Assume no volume change.

    This I have no idea how to do.
    Would appreciate it if someone could show me.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Undeadragons View Post
    1. Calculate the number of moles of nitrate ions in 2.20L of 2.02x10-3 mol L-1 lead(II) nitrate solution.
    Lead nitate has the formula \mbox{Pb(NO}_3)_2 thus every mol of lead nitrate contains two mols of nitrate ions (that is you get two nitrate ions for every molecule of lead nitrate).

    2.2 litres of 2.02\times 10^{-3} mol per litre solution contains  2.2 \times 2.02\times 10^{-3}=4.04 \times 10^{-3} mols of lead nitrate, and so 8.08\times 10^{-3} mols of nitrate ions.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,817
    Thanks
    316
    Awards
    1
    Quote Originally Posted by Undeadragons View Post
    2. Calculate the mass of solute which must be used in order to prepare the following solution 630 mL of 1.26 molL-1 KCl (aq) from KCl (s)
    The mass of 1 mol of KCl is
    (39.098~g) + (35.453~g) = 74.551~g

    So we want 630 mL of a 1.26 ml KCl(aq).

    \frac{0.630~\text{L soln}}{1} \cdot \frac{1.26~\text{mol KCl}}{1~\text{L soln}} \cdot \frac{74.551~\text{g KCl}}{1~\text{mol}} = 33.8163~\text{g KCl}

    Between my answer and CaptainBlack's, do you see what you need to do?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2007
    From
    Perth, Western Australia
    Posts
    4
    I think so, thanks.

    EDIT: Nope wait, i cant get number 4 or 6...
    Last edited by Undeadragons; July 25th 2007 at 05:00 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2007
    Posts
    4

    Lightbulb umm...

    Quote Originally Posted by topsquark View Post
    The mass of 1 mol of KCl is
    (39.098~g) + (35.453~g) = 74.551~g

    So we want 630 mL of a 1.26 ml KCl(aq).

    \frac{0.630~\text{L soln}}{1} \cdot \frac{1.26~\text{mol KCl}}{1~\text{L soln}} \cdot \frac{74.551~\text{g KCl}}{1~\text{mol}} = 33.8163~\text{g KCl}

    Between my answer and CaptainBlack's, do you see what you need to do?

    -Dan
    What have I done wrong. I did this and got a different answer:

    to work out the number of moles of KCl I thought it would be 1.26 x 0.63 = 0.7938
    then to work out how much KCl that is it would be 0.7938 x 74.551 = 59.1785838

    final answer I got was 59.1785838g of KCl is needed
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by DarkReviver View Post
    What have I done wrong. I did this and got a different answer:

    to work out the number of moles of KCl I thought it would be 1.26 x 0.63 = 0.7938
    then to work out how much KCl that is it would be 0.7938 x 74.551 = 59.1785838

    final answer I got was 59.1785838g of KCl is needed
    That looks OK to me, may be topsquark's arithmetic has gone wrong.

    RonL
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,817
    Thanks
    316
    Awards
    1
    Quote Originally Posted by DarkReviver View Post
    What have I done wrong. I did this and got a different answer:

    to work out the number of moles of KCl I thought it would be 1.26 x 0.63 = 0.7938
    then to work out how much KCl that is it would be 0.7938 x 74.551 = 59.1785838

    final answer I got was 59.1785838g of KCl is needed
    Apparently the gremlins got to my calculator again! Sorry 'bout that.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 4th 2011, 08:21 PM
  2. Replies: 6
    Last Post: July 26th 2010, 11:45 AM
  3. Chemistry
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: March 29th 2009, 05:58 PM
  4. Chemistry Stuff...
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: July 25th 2007, 06:05 AM
  5. Replies: 10
    Last Post: April 28th 2007, 04:00 PM

Search Tags


/mathhelpforum @mathhelpforum