# Math Help - Chemistry Stuff (Solutions)

1. ## 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.

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

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

4. I think so, thanks.

EDIT: Nope wait, i cant get number 4 or 6...

5. ## umm...

Originally Posted by topsquark
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

6. Originally Posted by DarkReviver
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

7. Originally Posted by DarkReviver
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