Making Molar Solutions From Solids

What are molar solutions? A molar solution is one that expresses “concentration” in moles per volume Usually the units are in mol/L mol/L can be abbreviated as M or [ ] Molar solutions are prepared using: a balance to weigh moles (as grams) a volumetric flask to measure litres L refers to entire volume, not water! Because the units are mol/L, we can use the equation M = n/L Alternatively, we can use the factor label method

Molarity Chemical representation of concentration M = moles of solute = mol liters of solution L Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1500 mL of solution.  What mass of NH4Cl is needed to prepare 450 mL of a 0.251 M NH4Cl solution?

Calculations with molar solutions Q: How many moles of NaCl are required to make 7.5 L of a 0.10 M solution? M=n/L, n = 0.10 M x 7.5 L = 0.75 mol # mol NaCl = 7.5 L x 0.10 mol NaCl 1 L = 0.75 mol But in the lab we weigh grams not moles, so … Q: How many grams of NaCl are required to make 7.5 L of a 0.10 M solution? # g NaCl = 7.5 L x 0.10 mol NaCl 1 L x 58.44 g NaCl 1 mol NaCl =43.83 g

Always start by labeling what you know Calculate the number of grams of AgCl formed when 0.400 L of 0.200 M Ag2SO4 reacts with MgCl2 Ag2SO4 (aq) + MgCl2 (aq)  MgSO4 (aq) + 2 AgCl(s) O.4 L ? g 0.2 M Think back to stoichiometry; what would you need to know in order to find grams of AgCl The answer is 22.88 g AgCl. Using the volume and molarity of Ag2SO4 calculate the moles. X moles/0.4 L = O.2 M  X=0.08 moles of Ag2SO4 Now you use stoich to convert moles of Ag2SO4 to grams of AgCl.

Now try this one … Consider the following equation: Na2SO4 + BaCl2  2 NaCl + BaSO4 What volume of a 0.15 M Na2SO4 will be needed to produce 57 g of BaSO4? The answer is 1.6 L of Na2SO4 If you want to solve for a volume of Na2SO4 and all you know is it’s molarity, then you are going to have to get the moles of Na2SO4. You can use stoich to convert grams of BaSO4 to moles of Na2SO4  57 g BaSO4 = 0.245 moles Na2SO4 So then 0.245 moles/ X L = 0.15 M

2Al(s) + 6 HCl(aq)  2 AlCl3(aq)+ 3H2(g) You have 275 mL of HCl and produce 2.1 g of H2. What was the original molarity of the HCl? You need to make 150 mL of a 0.5 M AlCl3, how many grams of Al should you start with? Answer to first problem = 7.6 M Answer to second problem = 2.03 grams Al Convert 2.1 g of H2 to moles of HCl  2.1 g H2 = 2.1 moles HCl, then use moles and L to calculate molarity 7.1 moles/0.275 L = 7.6 M 2nd problem First solve for moles of AlCl3  X moles /0.150 L = 0.5 M. x = 0.075 moles. Then convert 0.075 moles of AlCl3 to grams of Al  2.03 grams

Practice making molar solutions Calculate # of grams required to make 100 mL of a 0.10 M solution of NaOH (see above). Get volumetric flask, plastic bottle, 100 mL beaker, eyedropper. Rinse all with tap water. Fill a beaker with distilled water. Pour 20 - 30 mL of H2O from beaker into flask. Weigh NaOH. Add it to flask. Do step 5 quickly. Mix (by swirling) until the NaOH is dissolved. Add distilled H2O to just below the colored line. Add distilled H2O to the line using eyedropper. Place solution in a bottle. Place label (tape) on bottle (name, date, chemical, molarity). Place bottle at front. Rinse & return equipment.

More Practice Questions For more lessons, visit www.chalkbored.com How many grams of nitric acid are present in 1.0 L of a 1.0 M HNO3 solution? 2. Calculate the number of grams needed to produce 1.00 L of these solutions: a) 1.00 M KNO3 b) 1.85 M H2SO4 c) 0.67 M KClO3 3. Calculate the # of grams needed to produce each: a) 0.20 L of 1.5 M KCl b) 0.160 L of 0.300 M HCl c) 0.20 L of 0.09 mol/L AgNO3 d) 250 mL of 3.1 mol/L BaCl2 4. Give the molarity of a solution containing 10 g of each solute in 2.5 L of solution: a)H2SO4 b)Ca(OH)2 63 g 101 g 181 g 82 g a) 22 g b) 1.75 g c) 3 g d) 0.16 kg 5. Describe how 100 mL of a 0.10 mol/L NaOH solution would be made. a) 0.041 mol/L b) 0.054 mol/L