1. Dilutions

2. A dilution occurs when water is added to a concentrated solution. Because only water is being added, the moles of solute is unchanged. When the water is added, the concentration (molarity) of the solution will be reduced. Molarity = moles (stays the same) litres (gets larger) Mathematically, the molarity must get smaller. The moles are diluted.

3. What is the new concentration if 316cm 3 of a 0.417M solution of sulfuric acid is diluted up to a total volume of 906ml? M = moles litres Before: 0.417 M = moles 0.316L moles = 0.132 mol After: M = 0.132 mol 0.906L M = 0.146 M

4. The ratio of molarity and volume is the same before and after, because moles stays the same. Therefore, M = moles litres Could be written: M x litres = moles

5. If moles stays the same: before after M x litres = M x litres But we write it MV = M’V’ (.417)(.316) = M’(.906) M = 0.145M

6. Calculate the molarity of 0.856M of sulfuric acid in 450.0cm 3 of water that has 100.0cm 3 of water added to it. MV=M’V’ (0.856M)(0.450L) = M(0.550L) M = 0.700M

7. How much water must be added to make a 0.107M solution of nitric acid if the starting solution is 84.6ml of 0.932M nitric acid? MV = M’V’ (0.107)V = (0.0847)(0.932) V = 0.737L Total volume – Original volume = Must add 0.737L – 0.0847L = 0.652L