how to determine molar concentration and mass
- What is the molar concentration of an NaOH (aq) solution that contains 12.0 grams of NaOH (s) in 600. mL of solution?
- What mass of K3PO4 (s) is required to prepare 1500. mL of 0.20 M solution?
- Describe how you would prepare 250. mL of 0.100 M NaCl (aq). Make sure you show the calculation of the mass of the NaCl (s) that is required and a description of the process of making the solution.
- What volume of 0.020 M KOH (aq) solution contains 11.2 grams of KOH (s)?
- What is the resulting concentration of KBr (aq) when 100. mL of water are added to 400. mL of 0.350 M KBr (aq)? Assume the volumes are additive.
- How much water must be added to 500. mL of 1.00 M NaCl (aq) in order to make the final [NaCl (aq)] be 0.750 M?
- Given the equation:
2 NaI (aq) + Cl2 (g) 2 NaCl (aq) + I2 (s)
What volume of 0.100 M NaI (aq) must completely react in order to produce 508 grams of I2 (s)?
2 Answers | Add Yours
#1. Molar concentration of an NaOH solution. Molarity is always based on # of moles per 1000 mL of solution. In this case you have 12 g in 600 mL, therefore you have 20 g/L. @0g NaOH / 40 g/mole = 0.50 M solution.
#2. Mass of K3PO4 required. Molar mass is 212.27 g/mole. For one liter of 0.2 M solution you need 212.27*.2 = 42.45 g. For 1.5 L you would then need 42.45 * 1.5 = 63.68 g of K3PO4
#3. How prepare solution of NaCl. 0.1 M means you have 0.1M x 58.5 g/mole, or 5.85 g/1000 mL. To make 250 mL of solution, you need 5.85/4 = 1.46 g of NaCl. Weigh out 1.46 g and dissolve in 250 mL of water.
#4. Volume of 0.020 M KOH. A 1.0 M solution has 56.18 g/liter. Therefore a 0.02 M solution has 56.18 * 0.02 = 1.12 g/1000 mL or 11.2 g in 10 L.
#5. Resulting concentration. KBr = 119 g/mole. In 400 mL of 0.35 M you have 119 * .4 * .35 = 16.66 g. 16.66 g/500 mL = 33.32 g/L. 33.32/119 = 0.28 M.
#6. How much water add to make 0.75 M solution. 500/0.75 = 666.67 mL total volume. You already have 500 mL so have to add 166.67 mL
#7. Given the equation. molar mass of I2 = 253.8. Molar mass of NaI = 149.9. 508 g of I2 = 508/253.8 = ~2.0 moles. From your equation, for every 2 moles of NaI used, you produce 1 mole of I2, therefore you need 4 moles of NaI. 4 * 149.9 = ~600 g. A 0.1 M solution of NaI has 15 g/L. 600/15 = 40, so you need 40 L of the 0.1 M solution.
Mass of the empty dry beaker = 105.52 g
Initial mass of the beaker + CuCl2 (s) = 113.02 g
Initial mass of the two iron nails = 3.82 g
Final mass of the dry beaker and copper = 106.41 g
Final mass of the two iron nails = 3.04 g
We’ve answered 318,988 questions. We can answer yours, too.Ask a question