Question

Вычислите сколько миллилитров раствора KCl с массовой долей KCl 20%(плотность Р=1,13 г\мл) потребуется для приготовления 0,05 моль\л раствора хлорида каля объемом 10 л....

Answer 1

1) 1моль KCL=74.5*0.05=3.722
2)3.722*10=37.22

3)В 1 мл---1,13 мл
в 1000--37,22
х=1130/37.22=30.77мл.

4) 20%--30,37мл
   100%----х
х=30,37*100/20=151,85 мл калий хлорида

Answer 2

Calculating the Volume of KCl Solution Needed

To calculate the volume of the KCl solution needed to prepare a 0.05 mol/L potassium chloride solution with a volume of 10 L, we can use the formula:

Volume (in liters) = Number of moles / Molarity

First, we need to calculate the number of moles of KCl required for the 10 L solution.

The molarity of the solution is given as 0.05 mol/L, and the volume of the solution is 10 L.

So, the number of moles of KCl required is: Number of moles = Molarity × Volume

Number of moles = 0.05 mol/L × 10 L Number of moles = 0.5 moles

Now, we need to convert the moles of KCl to grams using the molecular weight of KCl.

The molecular weight of KCl is approximately 74.55 g/mol.

So, the mass of KCl required is: Mass = Number of moles × Molecular weight

Mass = 0.5 moles × 74.55 g/mol Mass = 37.275 grams

Now, we can calculate the volume of the 20% KCl solution needed to obtain 37.275 grams of KCl.

The density of the 20% KCl solution is given as 1.13 g/mL, and the mass of KCl required is 37.275 grams.

So, the volume of the 20% KCl solution needed is: Volume = Mass / Density

Volume = 37.275 g / 1.13 g/mL Volume ≈ 33.00 mL

Therefore, approximately 33.00 milliliters of the 20% KCl solution with a mass fraction of KCl at 20% and a density of 1.13 g/mL will be needed to prepare a 0.05 mol/L potassium chloride solution with a volume of 10 L.