# Re: Chemistry Questions! Any advice or help would be greatly appreciated!!!

Submitted by jusme on Tue, 10/20/2009 - 11:43

Ok so I started a Fundamentals of Chemistry course this past Thursday and the professor gave us 25 question and of the 25 Im having trouble with 7. Please any help would be greatly appreciated or any rreference tool or guidance.

1. Determine the mass of KCLO3 needed to prepare 875ml of a 1.80m solution.

2. Calculate the total # of moles of CL Ions for 0.380mgcl2 and 0.750m NACL in a 156ml solution.

3. Determine the concentration of 150.0 ml of a H3PO1 solution that required 72.65ml of 350m NA08 to neutralize.

4. Determine the concentration of CL Ions present when 14.0 millileter og 0.100m in a CL and 20.0 ml of 06.065mgcl2 solutions are combined.

5. Determine the mass of KO present in 300.0ml solutions that required 35.60 ml of 5.200m HC1 to neutralize.

6. Calculate the volume of 2.30 mncl2 needed to prepare 250.0 ml in a solution with a concentration of 0.30m CL Ions.

7. Determine the concentration of 300.ml of HC1 solution that required 15.4 ml in a 60.9 m NAOH to neutralize.

1.
To find the mass of a number of moles of a substance you need to calculate the molar mass of the substance

eg how many grams of sulphuric acid needed to make 400 mL of 0.02M solution.

Take atomic masses H=1, S=32 and O=16 (you use the values your course requires)
H2SO4 = 1x2 + 32 + 16x4 = 98g/mol

2.
To find the number of moles in a solution, you need:
No. of moles = volume in Litres x concentration in mole/L
= 0.400 x 0.02
= 0.008 moles H2SO4

1 mole weighs 98g, so
0.008 moles weighs (98/1) x 0.008 = 0.784g H2SO4

3.
If 25.00 mLof this solution required 35.50 mL of sodium hydroxide solution to neutralise it, what is the molarity of the NaOH?

No. of moles H2SO4 = 0.025 x 0.020 = 0.0005 moles.

H2SO4    +     2NaOH    ------>   Na2SO4    +    2H2O
1 mol             2mol
0.0005 mol    2 x 0.0005 = 0.0010 mole NaOH.

So molarity of NaOH = no.of moles/volume = 0.0010/0.0350 = 0.0286M

4.
What is the concentration of SO4 2- ions if 50 mL of 0.10 M H2SO4 is mixed with 250 ml of 0.04M Al2(SO4)3 solution?

Moles H2SO4 = 0.050 x 0.10 = 0.005 moles. As each mole of H2SO4 contain 1 mole of SO4 2- ions there will be 0.005 moles of SO4 ions

Moles Al2(SO4)3 = 0.250 x 0.04 = 0.01 moles. As each mole of Al2(SO4)3 contains 3 moles of SO4 ions, 0.01 moles will contain 3 x 0.01 = 0.03 moles of SO4 ions.

Total SO4 ions = 0.005 + 0.03 = 0.035 moles. Total volume of solution = 300 mL, so molarity of SO4 ions will be 0.035/0.300 = 0.167 M

This should point you in the right direction for some of these calculations

Oh great, thank you so much! This is very helpful.