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 K
_{3}PO_{4 (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)}+ Cl_{2 (g)}2 NaCl_{(aq)}+ I_{2 (s)}

What volume of 0.100 M NaI_{(aq)}must completely react in order to produce 508 grams of I_{2 (s)}?

### 2 Answers | Add Yours

Given that

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

#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.

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