Question

Какая масса хлорида цинка потребуется для приготовления раствора этой соли объёмом 500 мл. и с концентрацией 1.15 моль/л А)78 Б)79 В)77 Г)76 Д)75

Answer 1

Ответ:

А) 78

Объяснение:

Answer 2

Calculating the Mass of Zinc Chloride Required

To calculate the mass of zinc chloride required to prepare a solution with a volume of 500 ml and a concentration of 1.15 mol/L, we can use the formula:

Concentration (mol/L) = moles of solute / volume of solution (L)

Rearranging the formula to solve for moles of solute gives us:

Moles of solute = Concentration (mol/L) × volume of solution (L)

Then, we can use the molar mass of zinc chloride to convert moles to grams.

Calculation

Given: - Volume of solution (V) = 500 ml = 0.5 L - Concentration (C) = 1.15 mol/L

Using the formula: Moles of solute = Concentration × Volume of solution

Moles of solute = 1.15 mol/L × 0.5 L = 0.575 moles

The molar mass of zinc chloride (ZnCl2) can be calculated as follows: - Molar mass of Zn = 65.38 g/mol - Molar mass of Cl = 35.45 g/mol - Molar mass of ZnCl2 = 65.38 g/mol + 2(35.45 g/mol) = 136.28 g/mol

Now, to find the mass of zinc chloride required: Mass = Moles × Molar mass = 0.575 moles × 136.28 g/mol = 78.31 grams

So, the mass of zinc chloride required for the solution is approximately 78.31 grams.

Answer

The closest option to the calculated mass of zinc chloride (78.31 grams) is A) 78.