KONSENTRASI LARUTAN : MOLARITAS, MOLALITAS, DAN FRAKSI MOL

Chemical reactions in the laboratory or in everyday life usually occur in the form of a solution which is a homogeneous mixture of dissolved and dissolved substances with a certain composition. The composition of each solution depends on the ratio of the amount of dissolved substances to the solution, which is then called concentration.

Quantitatively, the concentration of solubility is stated in various units, including molarity, molality, and fractimol. Let's discuss one by one. Molarity or molarity, which is symbolized by the letter M capital, is the number of mols dissolved in 1 liter of solubility.

mathematically it can be written that the same amount of mol is dissolved divided by the dissolved volume in liters so that the units are mol per liter or molar Next, because mol is the same as mass divided by the mass of the relative molecule or MR and the volume can be converted to milliliters then this similarity can be reduced to

The same amount of molarity as the dissolved matter divided by the amount of the relative molecule or MR multiplied by 1000 divided by the volume in milliliters. Next, besides using these two equations, we can also calculate the molarity of a solution with the equation

The same amount of molarity as the melting type multiplied by the weight percentage multiplied by 10 divided by the melting relative mass or MRZ. Let's continue to the examples of the questions. The first question, determine the molarity of 0.05 mol of crystal MgSO4 that is dissolved in 500 ml of solution.

For this question, the molarity is determined by the similarity, the same as molmgSO4 divided by the solution volume in liters. Since the solution volume is known to be 500 ml, then we convert it first to liters, which is 0.5 liters.

Next in the question, it is known that the molmgSO4 is 0.05 mol, so if we substitute the existing equivalence, the molarity is equal to 0.05 mol divided by 0.5 liters, the result is 0.1 mol/liter or 0.1 molar.

The second question, how much is the solution that is made by dissolving 5.85 grams of NaCl in 500 ml of water? For this question, we first determine the relative mass of the NaCl, where the relative mass is the relative mass of sodium plus the relative mass of chlorine.

Well, since there is already a relative atom mass data, let's just substitute it, then the relative mass of the molecule is equal to 23 grams per mole plus 35.5 grams per mole, the result is 58.5 grams per mole.

Next we calculate the molarity with the same, the same molarity as NaCl divided by the relative molecule or MR times 1000 divided by the solution volume in milliliters.

The question is known that the NACL mass is 5.85 grams and the solution volume is 500 ml. So the molarity is the same as 5.85 grams divided by 58.5 grams per mole multiplied by 1000 divided by 500 liters minus 1. The result is 0.2 moles per liter or 0.2 molar. The third question

The solution of sulfide compounds generally contains solid nitrate acid that contains 63% HNO3. If the solution mass is 1.63 g/ml, then the molarity of the solid nitrate acid is

For this question, the molarity is determined by using the similarity mol = mass of the solution multiplied by the weight percent multiplied by 10 divided by the mass of the relative molecule or MR In the question, the mass data of the weight percent and the mass of the relative molecule or MR of HNO3

we immediately substitute so that the molarity is equal to 1.63 g/ml multiplied by 63 multiplied by 10 divided by 63 g/mol the result is 16.3 molars molality or molality which is symbolized by the letter m is the number of mols dissolved in 1 kg of solution

mathematically we can write molality is equal to the dissolved molecule divided by the dissolved mass in kilograms so the unit for molality is mol per kilogram or molal next because mol is equal to mass divided by the relative molecule or mr and the dissolved mass can be converted to grams then we can reduce this equation to

Molality is the same as the dissolved matter divided by the relative molecule or the dissolved matter of the dissolved matter multiplied by 1000 divided by the dissolved matter in grams. Let's continue with the example of the question. The first question, determine the molality of 0.2 mol BaCl2 that is dissolved in 500 grams of water.

For this question, the molality is determined by the similarity molality is equal to mol BaCl2 divided by the dilution time, which is water, in kilograms. In the question, the water is known to be 500 grams, then we first convert it to kilograms, so it becomes 0.5 kilograms.

Next in the question, it is known that the mol of BACL2 is 0.2 mol, so if we substitute the existing equivalence, molality is equal to 0.2 mol divided by 0.5 kg, the result is 0.4 mol per kg or 0.4 molal.

The second question, how much is the solution of 6 grams of glucose dissolved in 100 grams of water? For this question, we calculate the molality with the combination of molality = glucose mass divided by the relative molecule or MR glucose multiplied by 1000 divided by water mass in grams.

Well, the question is already known the mass data and the relative molecular mass or MR glucose as well as the water mass. Let's just substitute it, then the molality is equal to 6 grams divided by 180 grams per mole multiplied by 1000 divided by 100 kilograms minus 1. The result is 0.3 mol per kilogram or 0.3 molal.

The third question, the molality of 10% water mass in NaCl solution is? Known as the relative molecule or NaCl is 58.5 grams per mole. Well, if the percentage of NaCl solution is 10%, it means that in 100 grams of solution, the NaCl mass is 10 grams and the water mass is 90 grams.

And just like the previous question, the molality we determine by the similarity, molality is equal to the mass NaCl divided by the mass of the relative molecule NaCl multiplied by 1000 divided by the mass of water in grams.

Now we substitute all the data that is already there, then the molality is equal to 10 grams divided by 58.5 grams per mole multiplied by 1000 divided by 90 kilograms of minus 1. The result is 1.9 moles per kilogram or 1.9 molal.

The third concentration of solution is the mole fraction, which is the ratio of the number of moles of a substance to the number of total moles of substances in the solution. We symbolize this mole fraction with the capital letter X, which we can mathematically write: mole fraction ZA = mole ZA divided by the total mole of all substances in the solution.

Next, what you need to remember is the number of fractions of all components in the solution equal to 1. Now, to understand, let's continue to the example of the question. A solution is made by mixing 46 grams of ethanol and 44 grams of water. How many fractions of each substance in the solution?

It is known that the relative mass or MR ethanol is 46 grams per mole and the relative mass or MR water is 18 grams per mole. Well, we first determine the mole fraction for the ethanol with the combination of the mole fraction of ethanol is equal to the mole of ethanol divided by the mole of ethanol plus the mole of water.

Next, remember that the formula mol is mass divided by the relative mass of the molecule, then mol is the mass divided by the relative mass of the molecule, as well as mol .

The data for ethanol and water is already known in the question as well as the relative mass or MR of the two. Well, let's just substitute it, then the fraction of ethanol moles is equal to the ethanol mole, which is 46 grams divided by 46 grams per mole divided by 46 grams divided by 46 grams per mole plus the water mole, which is 44 grams divided by 18 grams per mole.

then the mole ethanol fraction is equal to 1 divided by 1 plus 2.44 equal to 1 divided by 3.44 the result is 0.29

Next, remember that the number of mol fractions of all components in the mixture is 1, then the ethanol fraction is added to the mol fraction of water equal to 1, so the mol fraction of water is equal to 1 minus the mol fraction of ethanol, equal to 1 minus 0.29, the result is 0.71.