Question

Сколько граммов карбоната калия потребуется для приготовления 2 литров 50%-го раствора (р=1,395г/см3)?

Answer 1

Ответ:

Объяснение:

{ρ1*V1*ω1 + ρ2*V2*ω2 = ρ3*V3*ω3;

{V1 + V2 = V3;

{1,51*V1*0,5 + 1,1*V2*0,1 = 1,19*1*0,2;

{V1 + V2 = 1;

{0,755V1 + 0,110V2 = 0,238;

{V1 + V2 = 1;

Answer 2

Calculating the amount of potassium carbonate needed

To calculate the amount of potassium carbonate (K2CO3) needed to make a 50% solution of 2 liters, we first need to determine the mass of K2CO3 in the final solution.

A 50% solution means that half the mass of the final solution will be K2CO3. The total mass of the final solution is 2 liters * 1.395 g/cm3 = 2790 g. Half of this is 1395 g of K2CO3.

The molar mass of K2CO3 is 138.205 g/mol. To get 1395 g of K2CO3, we need 1395 g / 138.205 g/mol = 101 mol of K2CO3.

The formula for calculating the amount of solute needed is:

Amount (mol) = Concentration (mol/L) × Volume (L)

Plugging in the values:

Amount (mol K2CO3) = 101 mol Concentration (mol/L K2CO3) = 101 mol / 2 L = 50.5 mol/L Volume (L solution) = 2 L

Therefore, the amount of potassium carbonate needed is 101 mol × 138.205 g/mol = 1400 grams.