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Question 26 (2 points} What is the molarity of solution that is prepared by dissolving 5.467 sodium chloride in 150.0 mL of water?...

Question

Question 26 (2 points} What is the molarity of solution that is prepared by dissolving 5.467 sodium chloride in 150.0 mL of water?

Question 26 (2 points} What is the molarity of solution that is prepared by dissolving 5.467 sodium chloride in 150.0 mL of water?



Answers

What is the molarity of a solution prepared by dissolving $4.49 \mathrm{~g}$ of sodium chloride in water and diluting to $40.0 \mathrm{~mL}$ of solution?

You take 4.49 g Of sodium chloride, divided by its smaller mass, which is 54 4 g. That will convert it to moles, then divide that by .045 m. You'll obtain 1.71 more clarity, Yeah.

So if we want to find the total polarity we can find the moles in each solution and divide by the total volume. Um so to get the malls we'll take the volume and will multiply it by the polarity. The volume should be in leaders. So we divide 5000 when you multiply it by the polarity which is most per litre. Then the leaders will cancel out and you'll get moles and we'll do the same on solution number two, It's .0475. Leaders multiplied by the minority which is to a point 63 2 moles in one leader. Let's compute that. So .035 times .375 is zero point 0131. And then .0475 times .63 2.0 300 mm. Yeah. And then now we can find the morality of the overall solution. Let's change colors. Um It's too black again, it's the total moles over the total volume in leaders. So on the top will add zero 13, 1 plus point 0300. And on the bottom zero point two or 35. This zero point 0475 leaders. Yeah, So the top comes out to to a point 431 in the bottom Comes out to zero 08-5 mm and we'll divide them to get our polarity, which is zero point 5 to 2 Mueller. Mhm.

Hi everyone. So in this question they ask what is the concentration of N. S. Here? So given Yeah, concentration of Yeah the nation If penetration of 15 Mahler 15 ml of solution. Subpoena medals. So who are lim 15 a man that is called 0.015 L. Mhm. Yeah. Mhm. With zero point similarity of silver nitrate. My dad Is he and not three. That is .2 for you. 03 smaller Then one Limo Bob. Let's see n. That is equal to 20.22. That is equal to 0.0 202 lita. So yeah. Yeah first calculate yeah Number of moles of silver nitrate and that is equal to .2503 in 20.020 To and that is equal to 0.0050611 mall. And that physically go And now fairness. No concentration of Tunisia. Yeah. Seconds concentration of N. S E. S. Oh and I see that is equal to end up on we. Yeah. That is equal to 0.506 double one, divided by 0.01 for you. And that is equal to zero 3374 Moller. And therefore concentration of Yeah. Do not Initial, not physical to zero 337 formal. Thanks.

Hello. Today we'll be talking to you about Chapter 15 Western 120 which tells us that we have 25 when zero grams of sodium bioscience me n a S, c n and that gets dissolved in 500 No leaders of water which same A 0.5 liters of water and were asked to calculate the concentration at the end of Thea disillusion. And so, first, if we want concentration, which has three units of Moeller, she was moles. Our leader. We need to find the number of moles of sodium fire signing. Do that. We need the molecular weight of sodium fire signing which we confined by adding up the elements on the periodic table and getting 81 072 grams of sodium. Elias, I need thermal insert a convert to moles. Let's just box off our givens. We have 25.0 grams of 13 sides. I need sodium fire assigning excuse times. One mole divided by 81 went 072 grams. That's going to give us zero when 308 moles of N E s u. Next. Now that we have moles with moles with the leaders. You're all we need to do is just divide and figure out what. Our final concentrations. 0.308 moles, 30 times sodium fire Signing time. Question Fighting by zero point size leaders will give us the final concentration. Zero Wait 616 bowls. Sorry, Moeller, Uh, in a sec. So hopefully this is helped you become a little bit more comfortable with the idea of converting from Graham Smalls and using that and volumes to determine concentration of solutions. Thanks. And here is our final


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