1. A solution is made by dissolving 13.5 g of glucose...

1. A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of water. What is the mass percentage of solute in this solution? 2. A commercial bleaching solution contains 3.62% by mass of sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2.50 kg of bleaching solution? 3. a) A solution is made by dissolving 4.35 g of glucose (C6H12O6) in 25.0 mL of water. Calculate the molarity of glucose in the solution. b) The density of water is 1.00 g/mL. Using density as a conversion from mL to mass (g), Calculate the molality of this glucose solution. 4. What is the mole fraction of HCl in a solution with 36 g HCl in 64 g H2O? 5. What is the molality and molarity of a solution made by dissolving 36.5 g of naphthalene (C10H8) in 425 g of toluene (C7H8)? 6. Commercial aqueous nitric acid has a molarity of 16 M. How many grams of HNO3 are in 12.5 mL of this solution? 7. Calculate the mass in grams of solute present in 600 mL of a 0.250 M SrBr2 solution. 8. Describe how I could make 500.0 mL of 0.75 M solution of HCl from a stock solution that is 12 M in concentration. 9. Calculate the molarity of 25.0 mL of 3.50 M HNO3 diluted to 0.250 L. 10. Ascorbic acid (Vitamin C, C6H8O6) is a water-soluble vitamin. A solution containing 80.5 g of ascorbic acid dissolved in 210 g of water has a density of 1.22 g/mL at 55 degrees Celsius. Calculate (a) the mass percent (b) the mole fraction (c) the molality (d) the molarity of ascorbic acid in this solution.

Transcript

00:03 So this first problem is a mass percent problem.

00:07 So in order to solve for mass percent, we need the mass of the solute over the mass of the solution.

00:15 We will then multiply our answer by 100%.

00:20 The solute in this problem is c6, h12.

00:26 Of 06 and we have 13 .5 grams of the solution or the solute excuse me so we know this is the solute because the solute is the substance that is being dissolved and in just a second we'll look at our solvent which is the substance that is doing the dissolving and in this problem and most problems the substance that is doing the dissolving is water and it says that we have 0 .10 .0 kilograms of h2o.

01:38 So the first problem here is that our salt you in our solvent do not have the same units.

01:47 So we'll need to do a conversion to either change our grams in our salt to kilograms or change our kilograms or so i'm going to go ahead and change my kilograms to grams in this problem.

02:07 So to do this conversion, we know that we have 1 ,000 grams and 1 kilogram.

02:20 So we'll multiply across the top, divide across the bottom, and we get 100 grams of solvent.

02:40 So now we can go ahead and just plug in our values into the formula above, so we'll go ahead and take our amount of solvent and put that on top.

02:54 So we have the 13 .5 grams of glucose.

03:08 And we're going to put that over our amount of solution.

03:12 So we can't just use the amount of our solvent.

03:15 We have to also take into consideration the solute and the solvent.

03:20 So we will add these two values together.

03:23 So our amount of solution is 130.

03:27 15 .5 grams, multiply all of this by 100%.

03:42 And our final answer is 11 .89%.

04:14 So in this problem, the question wants us to use the mass percent to calculate the mass of solute in solution.

04:26 So it says that we're starting with an amount of 2 .50 kilograms of solution and that 3 .62 % of that solution is sodium hypochlorite, which is the solute that we're solving for.

05:22 So we have 3 .62 % solute in this solution.

05:31 So in order to calculate our amount of solute by mass in the solution, we need to take the amount of solution, so that 2 .50 kilograms of solution and multiply by the percentage.

06:01 And when we multiply by the percentage, we have to move the decimal point in two places.

06:06 To the left giving us 0 .0362.

06:24 And when you do this calculation, you find that 0 .0905 kilograms of this solution is sodium hypochlorite or your solute.

07:00 Problem number three has two parts and the first part wants us to calculate the molarity of the solution, which is moles over a liter.

07:09 And the problem says we're starting with 4 .35 grams of glucose and we have 25 milliliters of solution.

07:18 So we need to do some conversions to start the problem.

07:22 So i'll start by converting my grams to moles.

07:26 I'm going to do this using the molar mass.

07:28 So i will change the grams to moles by putting one mole of glucose in the numerator, and i will use 180 grams of glucose as the denominator.

08:01 This gives me an answer of approximately 0 .04 moles of glucose.

08:15 Now i need to convert my milliliters of solution to liters, because that's the unit we need to work with this problem.

08:28 So to convert mill liters to liters, i'm just going to use the conversion factor because we know that in one liter, there are 1 ,000 millilators.

08:44 And when you do this calculation, so the answer to this conversion is 0 .1.

09:00 0 .025 liters.

09:11 And now that we have those conversions, we can solve the rest of the problem.

09:15 So we'll go ahead and take 0 .024 moles and divide by 0 .025 liters.

09:42 And this gives us an answer of approximately 0 .9 six, seven, molar.

10:07 So the second part of problem three wants us to calculate the molality.

10:11 So to calculate molality, we need to take moles over kilograms.

10:15 So we've already solved for our moles of glucose.

10:20 Now we need to convert our mill liters of solution to kilograms of solution.

10:27 So to do this, we're going to use the density given to us in the problem that one gram, the density of water is one gram per millilater.

10:44 So to convert our milliliters to grams, we'll use that density.

11:02 And then we need to convert from our grams to kilograms because the solution needs to be in kilograms.

11:16 And we know there's 1 ,000 grams in a kilogram.

11:21 So when we do this calculation, we get zero.

11:27 0 .025 kilogram.

11:40 So then we just need to plug in these values to calculate our molality.

11:46 So we will take again that 0 .04 moles divided by 0 .025 kilograms.

12:15 And we're going to get the same numeric answer this time, but the units do you need to be reported is those of molality.

12:25 So of 0 .967 moles per kilogram as our final answer for our molality of this solution.

12:53 The problem number four wants us to solve for our mole fraction...

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