Sifat Koligatif Larutan • Part 1: Konsentrasi (Molaritas, Molalitas, Fraksi Mol)

Hello everyone, welcome back to my channel, Christian Sutantio, on the channel Jendela Science. This channel is for those of you who want to understand mathematics, physics, and chemistry for high school. In this video, we will discuss the chemistry of high school students, namely the nature of the colligative solution. And in this first part, we will discuss the solution concentration.

For the other parts, you can click the playlist link on the top right. Or you can also click the link in the description to watch the video. Okay? To get a complete understanding, make sure to watch this video from start to finish. Okay, before we start, don't forget to click subscribe by pressing the button on the bottom right. And don't forget to press the bell so you don't miss our latest videos. Okay, let's get started.

The concept of solution. What is a solution? For example, you have a lot of salt, then you put it in a glass of water. After you put it in, you stir it until it dissolves. And what is formed? It is a solution of salt. From the example, the solution means it contains two components, which are dissolved and dissolved. The dissolved substance in the example is salt. The dissolves is water. Before we go deeper, I will briefly review some important formulas.

in this solution. The first one is m = n*mR. You have learned this in the 10th class of the concept of moles. m is mass, n is mole, and mR is the relative mass of molecules. The second one is rho = m/v. rho is mass, m is mass, and v is volume. Here, there is a formula mL = mT + mP.

L is the solution code, T is the solution code, P is the solution code. So, ML is the solution time, MT is the solution time, MP is the solution time. So, for example, the amount of salt is 10 grams, you put it in a glass of water that has a solution time of 100 grams, then if it is mixed, the solution time will be 100 plus 10 or 110 grams.

So this is like the law of mass loss. So the melting time is the number of the components. The melted time plus the melted time. Further on, there is V. V here is the volume. There are VL, VT, VP. The melting volume, the melted sat volume, and the melted sat volume. Remember earlier, rho equals m/v. So if we go back, what is V equal to? Then V is equal to m/rho.

and we go back again, what does M equal to? M equals Rho multiplied by V so here, if we know M, we have to find V, then it will be divided by Rho, the direction from M to V, divided from V to M, multiplied by V each component, if it's a solution, then it's a solution, so VL equals ML divided by Rho L ML equals Rho L multiplied by VL, understand? for the solution and the solution are the same but, this is what you need to pay attention to

VL is not equal to VT plus VP. So, what only occurs in the multiplication is the mass. The volume does not occur like this. So, if you are told that the volume is dissolved this much, the volume is dissolved this much, then the volume of the solution is not the multiplication. The volume is not multiplied. There is no such thing as the law of volume. Okay? Next, we go to mol or n. So, from mass to n is divided by mr. You have learned from class 10 about stoichiometry.

Because M is equal to N times MR, it means N is equal to M per MR. So if from mass to mole, it is divided by MR. Mass to mole, this is divided by MR. If T is T, if P is P. If from mole to mass, it means it is multiplied by MR. Now, it is also important to note that there is no such thing as NL, mole of solution. Why? Because this solution is a mixture of two different sets, so there is no MR.

the mass of the correlative mole, which every particle will have its own MR value. Salt has its own, water has its own. So if it's dissolved, salt water doesn't have MR, so there's no such thing as the "dissolved mole". Okay, you got it? Next, we will discuss the concentration of the dissolved.

The first is the molarity. Of course, you are already familiar with the term molarity. You have learned from the 10th and 11th grade. The molarity or the symbol M is n/V, the formula is mol divided by volume. Here we explain again, what is mol? Mol is dissolved, NT. And what is V? VL or dissolved volume. Mol is in mol and the dissolved volume is in liters.

for formula 1, it means this one, in liters. But in this case, most of the solution is in milliliters, so it's divided into this formula. So, NT is divided into m/mR. Remember, N = m/mR. So, NT = mT/mRT.

Then, the VL in milliliters must be divided by 1000. So, VL per 1000, or if multiplied by 1000 per VL, where the VL is in milliliters. So, this is the formula for instant, if the volume is in milliliters. Okay? The second one is similar to the first one. If it was molarity using R, now it's molality using L.

The symbol, if the molarity is M, if the molality is M, so here you don't get confused, okay? Earlier, M is also mass, okay? Here, M is also molality. So then, how to differentiate molality and mass?

In the previous slide, I have explained that there are ML, MT, and MP. So, if it's just M, it's a molality. If it's a matter of time, then what is the matter of time? If the matter is a solution, then the term is ML. If the matter is a solution, the term is MT. If the matter is a solution, the term is MP. So, it's clear that if M is a molality, then the matter must be given a code or index. Whether it's L, solution, T, solution, or P, solution. Okay? Molality is

The dissolved mol is divided by the dissolved mass in kilograms. Remember, the dissolved mass is in kilograms, the dissolved mol is mol. If we divide it, the dissolved mol is the same as the molarity, MT divided by MRT. But most of the time, the dissolved mass is in grams. So if you want to make it into kilograms, you have to divide by 1000 again. MP per 1000, so multiplied by 1000 per MP, where the MP is in? Okay? The third is the mol fraction.

What is a fractimol? A fraction means a part. For example, in this class there are 40 students. Out of these 40 students, there are 22 male students, and the rest are 18 female students. So, the fraction of male students is the total number of male students divided by the total number, which is 22 divided by 40.

and the fraction of women is the number of women divided by the total number 40. In the previous video, I have explained the concept of the solution.

The solution consists of two components, the dissolved matter and the dissolved matter. It means that if the dissolved mole fraction, Xt, is combined with the dissolved mole fraction, it is the dissolved mole divided by the total mole. The total mole means the dissolved mole plus the dissolved mole. It's almost the same as this. The number is 22 divided by 22 plus 18, 40. Okay? Now, Xp, the dissolved mole fraction means the dissolved mole divided by the dissolved mole plus the dissolved mole.

With that, we can conclude that Xt + Xp = Nt/Nt + Np + Np/Nt + Np. Since the names are the same, we can sum them up. So, we just need Nt + Np/Nt + Np, or equal to what? Equal to 1. So, Xt + Xp = 1.

Same as in the previous illustration, the male fraction 22/40 plus the female fraction 18/40 is 40/40 or 1. Okay, it's clear what the mole fraction is like. The fourth concentration of the solution is the mass percentage. What is the mass percentage? The symbol is m/m. The formula is mt/mL*100%. So here the denominator is always the solution that is always dissolved.

So, MT per ml, MT is the dissolved mass, and ml is the dissolved mass multiplied by 100%. Okay? The units are all in grams. Actually, it doesn't have to be in grams, it can also be in kilograms, kilograms, or milligrams, milligrams. But to make it easier to calculate, it's best to just divide it all by grams. The fifth is the percent of mass per volume. This is the percentage of m per v, the formula is MT per vr multiplied by 100%.

The principle is the same, the denominator is always dissolved, the denominator is always dissolved. So the dissolved mass in grams and the dissolved volume in milliliters. And the last one, the volume percentage. The percentage of V per V is Vt per VL times 100%.

Vt is the volume of the liquid dissolved in milliliters and Vl is the volume of the liquid dissolved in milliliters. This is also actually not binding, it can also be liters. But to make it easier in the calculation, we calculate if the volume is used in milliliters, if the mass is used in grams.

Next, let's go straight to the question. Determine the molality and the mol fraction of the solution made by adding 9 grams of glucose. Glucose is known as C6H12O6 in 900 ml of water. The water type is 1 gram per ml and is considered that the addition of glucose does not increase the volume of the solution.

Okay, we know that the water of C12, H1, and O16 is used to calculate the MR. So before we answer the question, we calculate the MR first. The MR of C6H12O6. Okay, the MR is the same as, remember, the method is 6 times the water of C, which is 12, plus 12 times the water of H, which is 1, plus

6 times the water O = 16. We calculate here 72 plus 12 plus 96. If we calculate, the total is 180. So the MR of glucose here is 180. Next, we calculate. What is the first one? The molarity. We use the instant method. The dissolved mass divided by the dissolved MR multiplied by 1000 per the dissolved volume.

Okay, so how much is the dissolved mass? The mass is glucose, here, 9 grams. The MR, we have already calculated it, 180 multiplied by 1000 per VL. What is VL? It's the dissolved volume. Here, it is considered that the addition of glucose does not increase the dissolved volume. It means that from 9 grams of glucose added to 900 ml of water, the dissolved volume remains the same. It is considered that it does not increase the volume. It means that we can directly add 900 ml of VL. Okay, here, there are two zeros, which can be corrected.

then 9 can be subtracted with this, there is 10/180, the zero can also be subtracted, then all of them are left 1/18 molar. Okay, now we go to the second one, namely molality. The formula is MT divided by MRT multiplied by 1000/MP, the melting time. So we put 9/180 multiplied by 1000,

per 900, why 900? well, here the water is 900 ml and the water type is 1 gram per ml, which means here the solution volume is 900 ml, then here the solution rho is 1, so if we calculate the solution time, m is equal to rho times v, the solution rho times the solution volume, which means 1 times 900, the result is 900 grams.

Okay, we put it here, which means 900. This is the same as M, we see the calculation is exactly 9/180 * 1000/900. This is also the same, so we immediately calculate it to be 1/18 molar. So it's a coincidence, this is not always, not always the molarity and the molarity are the same value, this is just a coincidence.

Okay, let's continue to the mole fraction here. Before we find the mole fraction, we must find the moles first. The mole is glucose, the mole is dissolved. That means the dissolved mass divided by the dissolved MR, or 9 divided by 180. The result is 0.05 moles. Then we calculate the mole of water, NP.

so the water mass is diluted, divided by the water MR the dilution mass is 900 grams from the calculation we put 900 here, then divided by the water MR, which is H2O, we calculate the MR of H2O here MRH2O, 2 times 1 plus 1 times 16, 2 plus 16, the result is 18

Okay, so 900 divided by 18, the result is 50. So, with that, we are asked to find the MOL fraction. If the command here is not known, then what is meant is the mixed MOL fraction, so the Xt that is being searched. Xt is equal to what? nT divided by nTOTAL or nT plus nP. So here, 0.05, 0.05 plus 50. Now, if we calculate, this will be 0.05

divided by 50.05 or if we divide it as a fraction, it becomes 1/1001 okay, the fraction of mol without units, yes, both Xt and Xp are without units okay, that's it for this video thank you for watching, if you like it, please like and share this video if you have any suggestions, criticisms, and input, you can write it in the comments section see you in the next video