Calculate the new molarity that results when 250. mL of...

Calculate the new molarity that results when 250. mL of water is added to each of the following solutions. a. 125 $\mathrm{mL}$ of 0.251 $\mathrm{MCl}$ b. 445 $\mathrm{mL}$ of 0.499 $\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ c. 5.25 $\mathrm{L}$ of 0.101 $\mathrm{M} \mathrm{HNO}_{3}$ d. 11.2 $\mathrm{mL}$ of 14.5 $\mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$

Calculate the new molarity that results when 250. mL of water is added to each of the following solutions.
a. 125 $\mathrm{mL}$ of 0.251 $\mathrm{MCl}$
b. 445 $\mathrm{mL}$ of 0.499 $\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$
c. 5.25 $\mathrm{L}$ of 0.101 $\mathrm{M} \mathrm{HNO}_{3}$
d. 11.2 $\mathrm{mL}$ of 14.5 $\mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$

Key Concepts

Molarity

Molarity is a measure of the concentration of a solution, defined as the number of moles of solute per liter of solution. It is a fundamental concept in chemistry used to express how much solute is dissolved in a given volume of solvent.

Dilution

Dilution refers to the process of reducing the concentration of a solution by adding more solvent without changing the number of moles of solute. This concept is key when calculating the new concentration after additional solvent is added to an existing solution.

Conservation of Moles

The principle of conservation of moles states that the number of moles of solute remains constant during the dilution process. This concept is used to determine the new molarity by dividing the original number of moles by the increased total volume of the solution.

Volume Conversion

Volume conversion is an important aspect of these calculations because molarity is defined in terms of liters. All volume measurements must be converted to liters to ensure proper application of the dilution equation and accurate results.

Transcript

00:01 Okay, this is from 31 from chapter 15 and it's a dilution question asking us what the new molarity is when 250 mil liters of water is added.

00:13 So the equation for this we're going to use is the molarity 1 or m1 v1 equals m2 v2 which is the first molarity times the first volume is equal to the second molarity and the second volume.

00:27 So for a you have a we have 125 milliliters of 0 .251 molar solution.

00:37 So m1, that is going to be the 0 .251m.

00:44 And v1 is going to be the bottom they gave us, so 125 milliliters.

00:51 We don't know what m2 is, so we're gonna put an x there.

00:58 We don't know yet, but v2 we do know because the question asks, with a new millerity on 250 milliliters or added so you just add 125 plus 250 and so the volume 2 is going to be 375 so when we write this out for the equation so we can set this up going to 251 times 125 milliliters is equal to 375 mill width.

01:57 And i'm going to put x here because we don't know what that is.

02:02 So basically what you do to get x is that you're multiplying the 0 .251 moles and the 125 milliers together.

02:10 And then you will divide those by 375.

02:19 And what that comes out to for your new molarity is going to be 0 .0837 molar.

02:39 Okay, so now we can go down to b, that's the first one.

02:47 So for b we have 445 molars of a .49 molar solution.

02:52 So once again we have m1 being the .449 molar solution.

02:53 So once again we have m1 being the .445.

02:57 99 molar b2 the 1 being the 445 milliliters you don't know what m2 is and b2 is 445 plus 250 it's going to be 600 and 95 milliliters okay we can now set up the equation again so it's going to be 4 4 4 millimeters times .49 molar that's going to be equal to 6995 x with x being m2 so do the so multiply these together so 445 times of 499 moles molars and i'm going to after you do that you're going to divide by the 695 which is going to come out to 0 .32.

04:29 Okay, we can go down to c now.

04:34 So see, we have 5 .25 liters of a .101 molar.

04:42 So m1 is a .101.

04:45 M2 is a .101.

04:47 M2 we they gave us leaders this time instead of milliliters...

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