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1 If 20.0 g of NaCl was brought to volume at 1.00 L, what % NaCl solution is this? (Remember 1ml of = 1g)2 How many milliliters of ethanol are in 1.5 liters of 20.0...

Question

1 If 20.0 g of NaCl was brought to volume at 1.00 L, what % NaCl solution is this? (Remember 1ml of = 1g)2 How many milliliters of ethanol are in 1.5 liters of 20.0% ethanol solution?3. Calculate and describe how you would prepare 500 mL of 3.ON HCI from 12.ON concentrated HCI:

1 If 20.0 g of NaCl was brought to volume at 1.00 L, what % NaCl solution is this? (Remember 1ml of = 1g) 2 How many milliliters of ethanol are in 1.5 liters of 20.0% ethanol solution? 3. Calculate and describe how you would prepare 500 mL of 3.ON HCI from 12.ON concentrated HCI:



Answers

For each of the following solutions, calculate the: a. grams of $2.0 \%(\mathrm{m} / \mathrm{m})$ NaCl solution that contains $7.50 \mathrm{g}$ of $\mathrm{NaCl}$ b. milliliters of $25 \%(\mathrm{m} / \mathrm{v})$ NaF solution that contains $4.0 \mathrm{g}$ of $\mathrm{NaF}$ c. milliliters of $8.0 \%$ (v/v) ethanol solution that contains $20.0 \mathrm{mL}$ of ethanol

First thing we will do here is to dissolve this compound in border will miss it, the conductors to see if the solution carries and electrical current. If solution is conducting, then we can determine whether the solution is a strong or weak electrolyte. By comparing its convicted, it's with that off unknown strong electrolyte, or it can also be done by using the APP critters shown in a textbook.

In this problem, we will first calculate the moles off each salute. Then we can calculate the volume in later from a polarity and the number off most off salute. Let's do the first one Well, we have to get the volume of the solution off an easy in. We first convert the mass off initial into molds off initial by using the conversion factor from mass to move the mamas off Initial is 58.44 g and that is why this conversion factor has been used here. It gives us as 0.3662 moles off any CIA. The solution volume will be more in salute diverted by polarity, which gives us the final solution as 0.1 thing. Six liters R 1 36 family off solution. Using the similar process for each other solution, we get the 4.3 g off C two at five wet as 62.2 amel solution and 0.85 g CST. Syria It solution as 47 Emel solution

Here we are looking to determine various different masses of solution depending on the quantity of sodium chloride that we desire. So in the first example we have the mass of the solid 10 g mass percent of any sales, 15%. The mass of the solution Is equal to 10 g, multiplied by 100 divided by 15 we get 67 g. In the first example, in the second example the mass The solution is equal to 25g, multiplied by 100, divided by 15 Is 167 grounds. In the next example, mask of solution is equal to 100g, Multiplied by 100 divided by 15. We get 667 g. And the final example in party the mass of solution is equal to 1.0 lb Most by by 100, divided by 15. We have £6.7.. Yeah.

Here we are converting the mass of sodium chloride to mass solution. Using 15% by mass as a conversion factor 15% by mass simply means we have 15 g of that salute for every 100 g of the solution. That ratio ends up being our conversion factor. So if we want 10 g of any CL and we know that we get 15 for every 100 then we need less than 100. This conversion factor 100 g for every 15 multiplied by the 10 gives us 66.7 g of solution. Next, we want 25 g and a C l. Multiply that by the 100 g solution to 15 g excl and we get 167 g of solution should be greater than 100 because we only get 15 for every 100 we want 25. Next. We want 100 g of n a C l. We get 15 for every 100 g of solution. So we're going to need in this case 667 g of solutions. Last we want a pound of any CL First we need to convert the pound into grams. Recognizing there's 453 6 g of N A c l in a pound of N A c l. Then we can use the 100 g solution to 15 g and a C L is a conversion factor and we get 3020 g of solution will contain £1 of N A. C L.


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