Question

Какой объем 1 М раствора хлорида хрома (3) необходимо взять для приготовления 200мл 2% раствора плотностью 1,018 г\мл

Answer 1

Масса сухого трихлорида хрома равна 200*0,02*1,018=4,072 г.
Количество вещества трихлорида хрома равно 4,072/158,5=0,0257 моль
1 л - 1моль
х л - 0,0257 моль
х=0,0257 л =25,7 мл
Ответ: 25,7 мл

Answer 2

Calculating the Volume of 1M Solution of Chromium Chloride (CrCl3)

To calculate the volume of a 1M solution of chromium chloride (CrCl3) needed to prepare a 200ml 2% solution with a density of 1.018 g/ml, we can use the formula:

Molarity (M) = moles of solute (n) / volume of solvent (V in liters)

First, we need to find the moles of solute required for the 200ml 2% solution. Then, we can use the molarity formula to calculate the volume of the 1M solution.

Finding Moles of Solute for 200ml 2% Solution

The 2% solution means there are 2g of solute in 100ml of solution. Therefore, for 200ml of solution, the mass of the solute (CrCl3) can be calculated as:

Mass of solute = (2g / 100ml) * 200ml = 4g

Calculating Moles of Solute

Using the molar mass of chromium chloride (CrCl3), which is approximately 158.36 g/mol, we can calculate the moles of solute:

Moles of solute = Mass of solute / Molar mass of CrCl3

Moles of solute = 4g / 158.36 g/mol ≈ 0.0253 moles

Calculating Volume of 1M Solution

Now, using the molarity formula, we can calculate the volume of the 1M solution needed:

1M = 0.0253 moles / V (in liters)

V (in liters) = 0.0253 moles / 1M

V (in liters) ≈ 0.0253 liters

Converting the volume to milliliters:

V (in ml) = 0.0253 liters * 1000 = 25.3 ml

Therefore, approximately 25.3 ml of the 1M solution of chromium chloride is needed to prepare the 200ml 2% solution with a density of 1.018 g/ml.