Question

Какие массы 15%-го раствора хлорида натрия и воды необходимы для приготовления 500 г 12%-го раствора NaCl?

Answer 1

W₀=0,15
m=500 г
w=0,12

масса соли
m(NaCl)=m₀w₀=mw

масса исходного раствора
m₀=mw/w₀

m₀=500 г*0,12/0,15=400 г

масса добавляемой воды
Δm(H₂O)=m-m₀

Δm(H₂O)=500 г - 400 г =100 г

Answer 2

Calculating the Masses of 15% Sodium Chloride Solution and Water

To calculate the masses of 15% sodium chloride solution and water needed to prepare 500g of a 12% sodium chloride solution, we can use the following formula:

C1V1 + C2V2 = C3V3

Where: - C1 = concentration of the first solution (15%) - V1 = volume of the first solution (unknown) - C2 = concentration of the second solution (0% for water) - V2 = volume of the second solution (unknown) - C3 = concentration of the final solution (12%) - V3 = total volume of the final solution (500g)

We can solve for V1 and V2 using this formula.

Calculating the Mass of 15% Sodium Chloride Solution

Using the formula C1V1 + C2V2 = C3V3, we can solve for V1:

15%V1 + 0%V2 = 12% * 500g

15%V1 = 12% * 500g

V1 = (12% * 500g) / 15%

V1 = 400g

So, 400g of the 15% sodium chloride solution is needed.

Calculating the Mass of Water

Using the formula C1V1 + C2V2 = C3V3, we can solve for V2:

15% * 400g + 0%V2 = 12% * 500g

0%V2 = 12% * 500g - 15% * 400g

V2 = (12% * 500g - 15% * 400g) / 0%

V2 = 100g

So, 100g of water is needed.

Therefore, to prepare 500g of a 12% sodium chloride solution, you would need 400g of 15% sodium chloride solution and 100g of water.