a sulfuric acid solution that is 65.0 h2so4 by mass has a density of 1.55 g/mL. What is the molality of sulfuric acid in this solution
1.55 g/mL x 1000 mL x 0.65 = 1007.5 g H2SO4.
mols H2SO4 = grams/molar mass
mass solution = 1.55 x 1000 = 1550.
mass H2O = 1550-1007 = ?
molality = mol H2SO4/kg H2O
To find the molality of sulfuric acid in the solution, we can use the formula:
Molality (m) = moles of solute / mass of solvent (in kg)
1. First, let's calculate the mass of sulfuric acid in the solution.
If the solution is 65.0% H2SO4 by mass, we can assume we have 100g of the solution.
So, the mass of H2SO4 in the solution is 65.0g (65.0% of 100g).
2. Next, we need to calculate the moles of H2SO4. To do this, we will use the molar mass of H2SO4, which is:
2*(1.01g/mol for H) + 32.07g/mol for S + 4*(16.00g/mol for O) = 98.09g/mol.
So, the number of moles of H2SO4 in the solution is:
moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4
moles of H2SO4 = 65.0g / 98.09g/mol = 0.66324 mol.
3. Finally, we need to calculate the mass of the solvent. We know that density (d) is defined as mass (m) divided by volume (v):
d = m / v
Rearranging the equation, we can find the mass (m) of the solvent:
m = d * v
Given that the density of the solution is 1.55 g/mL and assuming the volume (v) is 100 mL (since we assumed 100g of the solution), we can substitute the values:
m = 1.55 g/mL * 100 mL = 155 g.
Since the molality formula requires the mass of the solvent in kg, we need to convert the mass to kg by dividing by 1000:
mass of solvent (kg) = 155g / 1000 = 0.155 kg.
Now, plugging the values into the molality formula:
molality = moles of H2SO4 / mass of solvent (in kg)
molality = 0.66324 mol / 0.155 kg
molality = 4.28 mol/kg.
Therefore, the molality of sulfuric acid in this solution is 4.28 mol/kg.
To determine the molality of the sulfuric acid solution, we need to use the given information about the concentration and density.
Molality (m) is defined as the amount of solute (in moles) divided by the mass of the solvent (in kilograms). In this case, the solute is sulfuric acid (H2SO4) and the solvent is the solution itself (a mixture of sulfuric acid and water).
First, we need to calculate the mass of the sulfuric acid in the solution. We know that the solution is 65.0% H2SO4 by mass and that the density of the solution is 1.55 g/mL.
Let's assume we have 100 g of the solution (you can choose any mass to make calculations easier). This means we have 65.0 g of H2SO4 in the solution.
To find the volume of the solution, we can use the density formula:
Density = mass / volume
Rearranging the formula to isolate the volume:
Volume = mass / density
Using the values given:
Volume = 100 g / 1.55 g/mL
Volume = 64.52 mL
Now, we have the mass of H2SO4 (65.0 g) and the volume of the solution (64.52 mL).
To convert the volume from milliliters to liters, we divide by 1000:
Volume = 64.52 mL / 1000 mL/L
Volume = 0.06452 L
Now, we can calculate the molality using the equation:
Molality (m) = moles of solute / mass of solvent (in kg)
The molar mass of H2SO4 is:
2(1.008 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 98.09 g/mol
To calculate the moles of H2SO4, we use the mass and molar mass:
Moles = mass / molar mass
Moles = 65.0 g / 98.09 g/mol
Moles = 0.6629 mol
The mass of the solvent (water) is given by:
Mass of solvent = mass of solution - mass of solute
Mass of solvent = 100 g - 65.0 g
Mass of solvent = 35.0 g
Converting the mass of solvent to kilograms:
Mass of solvent = 35.0 g / 1000 g/kg
Mass of solvent = 0.035 kg
Now we can determine the molality:
Molality = moles of solute / mass of solvent (in kg)
Molality = 0.6629 mol / 0.035 kg
Molality = 18.940 mol/kg
Therefore, the molality of the sulfuric acid solution is approximately 18.940 mol/kg.