Kimia XII - Sifat Koligatif - Penurunan Tekanan Uap Larutan

Assalamualaikum Wr. Wb. Welcome back to the Teladan's Materi channel. On this occasion, we will discuss the 12th grade chemistry, which is related to the quality of the liquid. On this occasion, we will discuss the reduction of liquid pressure.

Before we go into the decrease in steam pressure, we must first understand the definition of steam pressure. So, steam pressure is the pressure generated by the amount of steam on the surface of the liquid in a cold condition. If illustrated here, for example, there is pure solvent or in this case it is a pure solvent, for example water.

If the water is left, then above it will be in a small amount, namely the water vapor. Then, if this water vapor is closed above it, then until one day it happens in a cold state, where the water vapor is at its maximum. When the vapor is in a cold state, then this can be calculated

how much value of the vapor pressure, which is generated from the vapor from the surface of the water here. But if into this pure solvent is added a solvent that is not easy to evaporate, or a solution with a non-volatile solute, or a solvent that is difficult to evaporate,

then this will reduce the amount of liquid steam in this day, for example, the amount of water steam that is on the surface of the water. Why can this happen? Because the water to evaporate is prevented by the dissolved particles that are symbolized as yellow in this day.

This will prevent the water from experiencing a steam, so that the steam here is getting smaller and causing the steam pressure to decrease. Or if defined in a simpler sentence, if water is flooded with a liquid that is difficult to evaporate,

then it will prevent the water to evaporate, so the amount of water vapor above the liquid will decrease. So this happens what is called vapor pressure decrease. Now to calculate how much the vapor pressure decrease, then the rule of Raoult can be used, where the vapor pressure of the solution is proportional to the fraction of the solution and the vapor pressure of the solution.

So here, symbol P indicates the pressure of the vapor solution and P0 is the pure vapor solution pressure, Xp is the melting fraccimol. Because the solution is compared to the solution, where the solution experiences a drop in the vapor pressure, then to calculate delta P, the pressure of the pure vapor solution is reduced by the pressure of the solution vapor. That's to calculate delta P. Or in other words,

The relation with the dissolved matter, then delta P can also be calculated by using the formula P0 multiplied by Xt. Xt here is the dissolved fraction, while Xp above is the dissolved fraction.

For example, there is 90 grams of glucose, or the formula is C6H12O6, here dissolved into 171 grams of water at 25 degrees Celsius. If the pressure of the water vapor at that temperature is 17 mmHg and the MR of C6H12O6 is 180, then determine the A, the pressure of the vapor, then the B is the drop in vapor pressure.

If we look at the definition in the question, the 90 grams here, it means that this is the dissolved gram. Because this is dissolved into 171 grams of water, so 171 is the dissolved gram. Who is the dissolves here? Water or H2O. How much water is the MR? The MR of water is 18.

The pure water pressure, which is 17 mmHg, is equal to P0, or the pure water pressure. Okay, let's calculate the water pressure. We can use this equation. P = P0 x X . Let's write the equation first. From here, P = P0 x X . If we divide it further,

So, P0 x P2 is the formula of the mole of the solvent divided by the total mole, which is NT + NP. Remember the mole concept that to calculate the mole or N, the formula is grams divided by mR.

Okay, let's continue, so we enter the data that is already there. What is the value of P0? The value of P0 is 17 multiplied by the solution mol, which means the solution mol is water, so 171 divided by the water mR is 18, per the solution mol is glucose, 90 grams, per 180,

plus the melting mole is 171 divided by 18 so from here we get 17 171 divided by 18 is equal to 9.5 times 0.5, 90 divided by 18 plus 9.5 so the result is equal to what? equal to 17 multiplied by

9.5 divided by 10, or the final result will be 16.15, the unit is mmHg. This is for the first one, so this already answers question A.

which is the pressure of the liquid vapor. But the question B is the decrease in the pressure of the liquid vapor. To answer B, we can use the equation delta P = P0 times XT, or if the liquid vapor P has been found, it means we can also use delta P = P0 - P. So we just use this formula which is simpler because P has been known. So we just answer the question B.

So, there, delta P = P0 - P, or P0 = 17 - P = 16.15, so the value is 0.85, and the unit is mmHg. This is the change in the pressure of the steam, or the decrease in the pressure of the steam.

Okay, this is a material related to the decrease in steam pressure. Hopefully the material I am presenting can be understood well and can be useful for all of you. And don't forget to like and subscribe to the Madri Pradhan channel for more development of the material I am presenting. Okay, I will end it first for the decrease in steam pressure. Assalamualaikum warahmatullahi wabarakatuh.