Question

Какую массу 24% раствора серной кислоты нужно смешать с 300 мл воды чтобы получить 18 % раствор?

Answer 1

Пусть масса вещества H2SO4 равна х.
тогда по формуле: w2=m(в-ва)/m(р-ра)
0,18=х/(х/0,24+300)
0,75х+54=х
0,25х=54
х=216 г
Тогда масса раствора H2SO4:
m(H2SO4)=m(в-ва)/w1=216/0.24=900 г

Answer 2

Это задача с когог класса-????

Answer 3

Calculating the Mass of 24% Sulfuric Acid Solution Needed

To calculate the mass of 24% sulfuric acid solution needed to mix with 300 ml of water to obtain an 18% solution, we can use the following steps:

1. Determine the Initial and Final Concentrations: - The initial concentration of the sulfuric acid solution is 24%. - The final concentration after mixing with water is 18%.

2. Set Up the Equation: - Let's assume the mass of the 24% sulfuric acid solution needed is 'm1' and the mass of water is 300 grams. - We can use the formula for mixing two solutions to obtain a desired concentration: ``` C1V1 + C2V2 = C3(V1 + V2) ``` Where: - C1 and V1 are the concentration and volume of the first solution (24% sulfuric acid solution and 'm1' grams, respectively). - C2 and V2 are the concentration and volume of the second solution (0% sulfuric acid solution and 300 grams, respectively). - C3 is the final concentration (18% sulfuric acid solution). - V1 + V2 is the total volume of the final solution.

3. Solve for 'm1': - We can solve for 'm1' using the given concentrations, volumes, and the formula.

Let's proceed with the calculation using the provided information.

Using the formula: ``` C1V1 + C2V2 = C3(V1 + V2) ``` Substituting the given values: ``` (24%)(m1) + (0%)(300) = (18%)(m1 + 300) ``` Solving for 'm1': ``` 0.24m1 = 0.18(m1 + 300) 0.24m1 = 0.18m1 + 54 0.06m1 = 54 m1 = 54 / 0.06 m1 = 900 grams ``` So, the mass of 24% sulfuric acid solution needed to mix with 300 ml of water to obtain an 18% solution is 900 grams.

This calculation is based on the provided concentration and volume information