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examples of molarity and molality

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Question:Molarity, Molality & Dilution Notes and Practice Answer the questions below on a separate sheet of paper. SHOW ALL WORK, including units!! Watch your significant digits and CIRCLE YOUR ANSWERS. Molarity Just a reminder, molarity is one of the many ways to measure concentration or the strength of a solution. When using molarity to measure concentration you must follow the formula below and then put a capital M at the end of your answer to let the world know you used the molarity formula. M = moles of solute Liters of solvent 1.Calculate the molarity of a solution which contains 0.40 mol of a substance dissolved in 1.6 L of a solution. 2.What is the molarity of a solution containing 325 g of NaCl dissolved in 750. mL of solution? 3.140 g of KCl is dissolved in 600. mL of water. What is the molarity? 4.724.4 g of ammonium phosphate in 4500 mL of alcohol. What is the molarity of the solution? 5.You are making 2.2 L of 3.1 M silver nitrate solution. Who many moles of solute are there? 6.How many grams of MgCl2 are needed to make 700.mL of a 1.4 M solution? 7.93.2 g of copper (II) sulfate is mixed into 290. mL of water. What is the molarity? Molality Molality is an additional way to measure the strength or concentration of a solution. It is abbreviated with a little m and is calculate only slightly differently than molarity. Here is the formula. m = moles of solute kg of solvent You will be given two mass measurements and you must decide which is the solute and which is the solvent. Look for the phrases dissolved in, placed in, or mixed with to identify the two parts. The solute comes before the phrase and the solvent comes after. Change the solute into moles (factor label) Change solvent into kg (KHDBdcm) Molality and molarity can be very close if water is the solvent. Example: 190 g of CuSO4 are placed in 3500 g of water. Determine the molality. Solute: 190 g CuSO4 1mole = 1.2 mole CuSO4 159.9 g Solvent: 3500 g = 3.5 kg water Molality = 1.2 moles = 0.30m 3.5 kg Mixed Problems Decide if the problem is molarity or molality so you know which formula to use 8.What mass of calcium hydroxide must dissolve in 850 mL of water to make a 2.4 M solution? 9.326g of C6H6 dissolve in 820. g of acetone. What is the molality? 10.What mass of glucose must dissolve in 400. g of ethanol to make a 1.6 m solution? 11.What volume of water must be added to 325 g of chromium (III) carbonate to make a 0.90 M solution? 12.What mass of ethanol is 360. g of sucrose dissolved in to make a 1.6 m solution? If the density of ethanol is 0.89 g/mL, determine the volume of ethanol used. 13.Determine the molar mass of a solute if 78 g of it are dissolved in 1800 mL of water to make a 0.25 M solution. Dilution Problems You can make less concentrated solutions by diluting. You must use the same volume units throughout each problem (mL or L). Purpose to make a lower concentrated solution from a more concentrated stock solution. Terminology: Stock Solution most concentrated or original solution. (before) (after) M1V1 = M2V2 CAUTION: V2 is to the total volume. The amount of stock solution plus the amount of water added. Example: How would you prepare 100. mL of 0.40 M MgSO4 from a stock solution of 2.0 M MgSO4? M1V1 = M2V2 (2.0 M) V1 = (.40M) (100mL) V1 = 20 mL of stock solution In other words, in order to make the solution 20 mL of the stock 2.0 M solution needs to be measured with a graduated cylinder. Then 80 mL of water are added to the stock to get a total volume of 100 mL (the V2) 1.To what volume should 11.9 mL of an 8.00 M acetic acid solution be diluted to obtain a final solution that is 1.50 M? 2.What volume of a 5.75 M formic acid solution should be used to prepare 2.0 L of a 1.0 M formic acid solution? 3.What is the molarity of a solution of ammonium chloride prepared by diluting 50.0 mL of 3.79 M ammonium chloride solution to 2.0 L? 4.A 25.0 mL sample of ammonium nitrate solution produces a 0.186 M solution when diluted with 50.0 mL of water. What is the molarity of the stock solution? 5.What volume of 8.0 M stock solution must be used to make 750 mL of a 0.20 M solution? 6.What volume of water must be added to 30.0 mL of 9.2 M stock solution to make a 1.2 M solution? 7.Determine the molarity of a stock solution if 81.0 mL of it were used to make 650 mL of 0.60 M solution. 8.250 mL of water are added to 35 mL of 4.2 M stock. What is the new concentration? 9.To what volume does 15 mL of a 4.9 M solution need to be diluted in order to obtain a final solution that is 0.80 M?

Answers:ur ans is correct.no need more explanation.

Question:What happens to molarity, molality, and concentration of a solution when more solute is added? What happens when more solvent is added?

Answers:When more solute is added M , m and concentration of the solution increase When more solvent is added M , m and concentration decrease

Question:1. A bottle of phosphoric acid is labeled "85.0% H3PO4 by mass; density = 1.680 g/cm3." Calculate the MOLARITY of phosphoric acid in the solution? 2. A bottle of phosphoric acid is labeled "85.0% H3PO4 by mass; density = 1.680 g/cm3." Calculate the MOLALITY of phosphoric acid in the solution?

Answers:formula: M=mol/L know: 1 cm^3=1 L step 1: assume % is g step 2: find mol, start with 85 g, go to mol step 3: find L, start with 85 g, go to cm^3, go to L step 4: use M eqn and solve formula: m=mol/kg know: 1 kg=1000 g step 1: find g solute, start with 85 g, go to mol, go to g step 2: g obtain in step 1 subtract from g sln density which gives you g solvent step 3: g to kg step 4: use m eqn and solve good luck.

Question:The electrolyte in a car battery is a 3.75 M sulfuric acid solution with a density of 1.23 g/mL. i'm completely lost on how to find mass percent, normality, and formality, and i still don't completely understand the difference between molarity and molality.

Answers:Molarity= moles of solute/L of solution Molality=moles of solute/kg of solvent So, first let's figure out the mass percent: (which would be grams of solute/grams of solution multiplied by 100) You have 3.75M sulfuric acid. M=Mol Sulfuric Acid/L solution To make life easier, let's assume you have 3.75mol sulfuric acid in 1L of solution, which would give us 3.75M To find the amount of grams of sulfuric acid that you have, multiply the moles by the molecular weight. So: 3.75 mol * 68.03 g (molecular weight): About 255 grams. To find out how many grams of solution, convert the amount of solution in liters to grams using the density. So, we assumed we had 1L of solution. Convert 1L to mL (because the density is in g/mL): 1L * 1000 mL/1L= 1000mL solution So density=mass/volume. Plug in the information: 1.23 g/mL = mass/1000mL 1.23g/mL * 1000mL=1230 grams solution. So, mass percent= (255 grams H2SO4/1230 grams solution)*100%=20.7% Now let's calculate the molality: Molality= moles of solute/kg of solvent So, we already assumed that there are 3.75 mol H2SO4 in the solution, so we have the moles solute. To determine the amount of solvent: -We know there are 255g solute in the solution and 1230 grams in total. And we know, Solute + Solvent = Solution. So plug in, and solve for the solvent: 255g + Solvent = 1230g Solvent=1230g-255g Solvent=975g Now, we have to convert that to kg by dividing by 1000 (because molality is in kg). So to recap: We have 3.75 mol H2SO4 and .975 kg Solvent. Thus, the molality of the solution is: mol solute/kg solvent= 3.75 mol/.975 kg= 3.85m Ok, so for normality: n mol equivalent of solute/L solution I found the following on the internet (because I find the concept difficult to explain: "Normality, N, is similar to molarity, moles of solute per liter of solution. However, instead of the entire solute, the normality is based on the number of moles of the active part of the solute, called a chemical equivalent. For an acid, the chemical equivalent is the number of moles of H+1 ion. For a base, the chemical equivalent is the number of moles of OH-1 ions. For an oxidation-reduction solution, the chemical equivalent is the number of moles of electrons transferred. The normality of hydrochloric acid, HCl, is the same as the molarity of hydrochloric acid, because there is one mole of H+1 ions for every one mole of hydrochloric acid. The normality of sulfuric acid, H2SO4, is twice the molarity because there are two moles of H+1 ions per mole of sulfuric acid. The advantage to using normality is that it gives an effective concentration (3M sulfuric acid is twice as acidic as 3M hydrochloric acid this is clear if they are labeled 6N and 3N, respectively)." Therefore, the normality for H2SO4 will be twice the molarity. 3.75*2= 7.50 N I don't remember how to calculate formality, but I think it is 3.75 F because H2SO4 is not an ionic compound. This paragraph from the website I cited helped: Formality, F, is the number of formula weight units of solute per liter of solution. Remember that one mole of a compound has a mass equal to the formula weight in grams. The number of formula weight units is equal to the number of moles for molecular substances. The purpose of formality is to distinguish the number of moles of a compound from the number of moles of ions in solutions of ionic compounds or weak electrolytes. I hope I've helped a little! Good luck!

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Chemistry: Molarity, Molality, Mole Fraction :Watch more free lectures and examples of Chemistry at www.educator.com Other subjects include Algebra, Trigonometry, Calculus, Biology, Physics, Statistics, and Computer Science. -All lectures are broken down by individual topics -No more wasted time -Just search and jump directly to the answer

Molarity, Molality, Mass Percent, and Mole Fraction :