Question

Водный раствор содержит 577 г серной кислоты в 1 литре. Плотность раствора 1,335 г/мл. Вычислите массовую долю, молярную концентрацию, молярную концентрацию эквивалента, моляльность и молярные доли серной кислоты и воды в полученном растворе. Ответ: 43,22%; 11,76 моль/л; 5,88 моль/л; 7,76 моль/кг; 0,123; 0,877. Помогите, пожалуйста, половину задачи решила, а дальше не сходиться с ответами или не так делаю... 1. Массовая доля 577г. / 1335г. * 100% = 43,22% 2. См= 43,22*1,335*10/38 = 5,88 3. Сн= 2*5,88= 11,76. 4. М= ? 5. ??? 6. ???

Answer 1

Cm = n/m(H₂O) = 5,88/(1335-577) = 7,76(моль/кг)

n(H₂O) = 758/18 = 42,11(моль/л)

∑n(H₂O + H₂SO₄) =5,88+42,11=47,99

Xj(H₂SO₄)=5,88/47,99=0,123

Xj(H₂O) = 1 - 0,123 = 0,877

Answer 2

Let's go through the calculations step by step to find the missing values:

1. Mass fraction (mass percent): You have already correctly calculated this: Mass of sulfuric acid / Total mass of the solution * 100% = 577g / 1335g * 100% ≈ 43.22%

2. Molarity (M): The molarity is the number of moles of solute (in this case, sulfuric acid) per liter of solution.

Molarity (M) = Moles of solute / Volume of solution (in liters)

To find the moles of sulfuric acid, we can use the mass given and its molar mass.

Molar mass of sulfuric acid (H2SO4) = 2 * atomic mass of hydrogen + atomic mass of sulfur + 4 * atomic mass of oxygen Molar mass of H2SO4 = 2 * 1.008 g/mol + 32.06 g/mol + 4 * 16.00 g/mol ≈ 98.09 g/mol

Now, calculate the moles of sulfuric acid:

Moles of H2SO4 = Mass of H2SO4 / Molar mass of H2SO4 Moles of H2SO4 = 577g / 98.09 g/mol ≈ 5.88 mol

Molarity (M) = 5.88 mol / 1 L ≈ 5.88 mol/L

So, the molarity of the sulfuric acid solution is approximately 5.88 mol/L.

3. Molarity of Equivalent (Cn): The molarity of equivalent is used in redox reactions and is defined as the number of moles of a substance that can donate or accept one mole of electrons.

For sulfuric acid (H2SO4), since it can donate two moles of H+ ions, its molarity of equivalent (Cn) will be half of its molarity.

Cn = Molarity (M) / 2 Cn = 5.88 mol/L / 2 ≈ 2.94 mol/L

So, the molarity of equivalent for sulfuric acid is approximately 2.94 mol/L.

4. Molality (m): Molality is defined as the number of moles of solute per kilogram of solvent.

Molality (m) = Moles of solute / Mass of solvent (in kilograms)

First, calculate the mass of water (solvent) in 1 liter of the solution: Density of the solution = Mass of solution / Volume of solution 1.335 g/mL = Mass of solution / 1 L Mass of solution = 1.335 g

Since we know the mass of sulfuric acid is 577g, the mass of water will be: Mass of water = Mass of solution - Mass of sulfuric acid Mass of water = 1.335 g - 577 g ≈ 758 g

Now, calculate the molality:

Molality (m) = 5.88 mol / 0.758 kg ≈ 7.76 mol/kg

So, the molality of the sulfuric acid solution is approximately 7.76 mol/kg.

5. Molar fraction (X) of sulfuric acid: The molar fraction of a component in a solution is the ratio of the moles of that component to the total moles of all components in the solution.

Molar fraction (X H2SO4) = Moles of H2SO4 / Total moles in the solution

We already know the moles of sulfuric acid (5.88 mol). To find the total moles in the solution, we can use the density to calculate the volume and then use the molarity.

Volume of the solution = Mass of solution / Density of the solution Volume of the solution = 1335 g / 1.335 g/mL ≈ 1000 mL = 1 L

Total moles in the solution = Molarity * Volume of the solution Total moles in the solution = 5.88 mol/L * 1 L = 5.88 mol

Now, calculate the molar fraction of sulfuric acid:

X H2SO4 = 5.88 mol / 5.88 mol ≈ 1

So, the molar fraction of sulfuric acid in the solution is approximately 1.

6. Molar fraction (X) of water: Since we have already calculated the molar fraction of sulfuric acid to be 1, and there are no other components present in the solution, the molar fraction of water can be calculated as:

X water = 1 - X H2SO4 = 1 - 1 = 0

So, the molar fraction of water in the solution is 0.

In summary, the results are:

1. Mass fraction of sulfuric acid: 43.22%
2. Molarity (M) of the solution: 5.88 mol/L
3. Molarity of Equivalent (Cn) of sulfuric acid: 2.94 mol/L
4. Molality (m) of the solution: 7.76 mol/kg
5. Molar fraction (X) of sulfuric acid: 1
6. Molar fraction (X) of water: 0