on dissolveing 2.34g of solute in 40g of benzene . the boiling point of solution was higher than that of benzene by 0.81k. kb value of benzene is 2.53 k.kg mol inverse . calculate the molar mass of the solute.
delta T = Kb*m
delta T = 0.81
Kb = 2.53
Solve for m = molality
Then m = grams/molar mass. You know m and you know grams, solve for molar mass.
Post your work if you get stuck.
To calculate the molar mass of the solute, we can use the formula:
∆Tb = Kb * m
where:
∆Tb = the change in boiling point
Kb = the molal boiling point elevation constant for the solvent (benzene), which is 2.53 K.kg mol⁻¹
m = molality of the solution
First, we need to calculate the molality (m) of the solution using the given mass of solute and solvent.
Molar mass of benzene (C₆H₆) = 78.11 g/mol
Mass of solute = 2.34 g
Mass of benzene solvent = 40 g
Moles of solute = mass / molar mass = 2.34 g / molar mass
Moles of benzene = mass / molar mass = 40 g / 78.11 g/mol
Molality (m) = moles of solute / mass of the solvent in kg
= (2.34 g / molar mass) / (40 g / 1000)
Now, we have the molality (m) and the change in boiling point (∆Tb = 0.81 K).
0.81 K = 2.53 K.kg mol⁻¹ * m
Solving for m:
m = 0.81 K / 2.53 K.kg mol⁻¹
Finally, we can use the molality to calculate the molar mass.
moles of solute = m * mass of solvent in kg
= (0.81 K / 2.53 K.kg mol⁻¹) * (40 g / 1000)
Molar mass of the solute = mass of solute / moles of solute
Substituting the values:
Molar mass of the solute = 2.34 g / [(0.81 K / 2.53 K.kg mol⁻¹) * (40 g / 1000)]
Simplifying, we get:
Molar mass of the solute ≈ 186.32 g/mol
Therefore, the molar mass of the solute is approximately 186.32 g/mol.
To calculate the molar mass of the solute, we need to use the equation for boiling point elevation:
ΔTb = Kbm
Where:
ΔTb = change in boiling point
Kb = boiling point elevation constant (molal boiling point constant)
m = molality of the solution
First, let's calculate the molality of the solution using the given information:
Molality (m) = moles of solute / mass of solvent in kg
Given:
Mass of solute (m1) = 2.34 g
Mass of solvent (m2) = 40 g
Molar mass of benzene = 78.11 g/mol
First, we need to convert the mass of solute to moles:
Moles of solute = mass of solute / molar mass of solute
Moles of solute = 2.34 g / molar mass of solute
Now, we need to convert the mass of the solvent to kg:
Mass of solvent (kg) = mass of solvent (g) / 1000
Mass of solvent (kg) = 40 g / 1000
Next, we can calculate the molality:
Molality (m) = moles of solute / mass of solvent (kg)
Molality (m) = (2.34 g / molar mass of solute) / (40 g / 1000)
Now we have the value of molality, which we will use in the next step.
Given:
ΔTb = 0.81 K
Kb = 2.53 K⋅kg⋅mol^(-1)
Rearranging the equation ΔTb = Kbm, we can solve for molar mass:
Molar mass of solute = ΔTb / (Kb × molality)
Molar mass of solute = 0.81 K / (2.53 K⋅kg⋅mol^(-1) × molality)
Substituting the value of molality, we can find the molar mass of the solute:
Molar mass of solute = 0.81 K / (2.53 K⋅kg⋅mol^(-1) × [(2.34 g / molar mass of solute) / (40 g / 1000)])
This equation is a little complex to solve algebraically, so we will use an iterative method to approximate the molar mass. We can start with an initial estimate, calculate the value on the right side, and then refine the estimate until we converge on a solution.
Let's choose an initial estimate for the molar mass, such as 100 g/mol:
Molar mass of solute = 0.81 K / (2.53 K⋅kg⋅mol^(-1) × [(2.34 g / 100 g/mol) / (40 g / 1000)])
Now we can calculate the molar mass using this estimate and repeat the calculation using the new molar mass value until the value converges. The final calculated value of the molar mass will be the answer to the problem.