# 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...

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)}?

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#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.

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

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