The freezing point of a 5% solution of boric acid is -1.55 degrees Celsius.How many grams of boric acid should be used in preparing 500mL isotonic solution?
To determine the number of grams of boric acid needed to prepare a 500mL isotonic solution, we need to use the freezing point depression formula:
ΔT = K × m
where ΔT is the change in freezing point, K is the cryoscopic constant for boric acid, and m is the molality of the solution.
From the given information, we know that the freezing point depression is -1.55 degrees Celsius and the solution is 5% boric acid. We need to calculate the molality of the solution.
Step 1: Calculate the molality (m)
Molality (m) = moles of solute / kilogram of solvent
Since we are preparing a 500mL solution, the mass of the solvent (water) is 500g or 0.5kg.
The molar mass of boric acid (H3BO3) is approximately 61.83 g/mol.
To calculate the moles of solute, we use the concentration and the molar mass of boric acid:
moles of solute = concentration × volume / molar mass
Concentration = 5% = 5 g/100 mL = 5 g/0.1 L = 50 g/L
moles of solute = (50 g/L) × (0.5 L) / (61.83 g/mol)
moles of solute ≈ 0.405 mol
Now, we can calculate the molality:
m = 0.405 mol / 0.5 kg
m ≈ 0.81 mol/kg
Step 2: Calculate the freezing point depression constant (K)
The cryoscopic constant (K) for boric acid is 2.11 °C/m.
Step 3: Calculate the change in freezing point (ΔT)
ΔT = K × m
ΔT = 2.11 °C/m × 0.81 mol/kg
ΔT ≈ 1.71 °C
Step 4: Calculate the amount of boric acid needed
The freezing point depression is the difference between the normal freezing point of pure water (0 °C) and the freezing point of the solution (-1.55 °C).
Therefore, the amount of boric acid needed can be calculated using the formula:
Amount of solute = molality × molar mass × kg of solvent
Amount of solute = 0.81 mol/kg × 61.83 g/mol × 0.5 kg
Amount of solute ≈ 24.95 g
Therefore, approximately 24.95 grams of boric acid should be used in preparing a 500mL isotonic solution.
To find out how many grams of boric acid are needed to prepare a 500 mL isotonic solution, we need to determine the concentration of the isotonic solution first.
An isotonic solution is a solution that has the same osmotic pressure as another solution. In this case, we want to prepare a 5% isotonic solution.
Since the freezing point of the 5% solution of boric acid is given as -1.55 degrees Celsius, we can use this information to calculate the molar concentration of boric acid.
The formula to calculate the molar concentration of a solute in a solution is:
molality (m) = moles of solute / mass of solvent in kg
First, we need to convert the freezing point depression (-1.55 degrees Celsius) to kelvin:
Tf = -1.55 + 273.15 = 271.60 K
Next, we need to calculate the molal concentration (m):
m = Kf x molar mass of boric acid / change in freezing point
Here, Kf is the cryoscopic constant, which for water is 1.86 °C kg/mol.
Let's assume the molar mass of boric acid (H3BO3) is M grams/mol, then we can rearrange the formula to solve for m:
m = 1.86 / (Tf * M)
Since we want to prepare a 5% solution, we know that the molality (m) will be 0.05 mol/kg.
Setting the equation equal to 0.05, we have:
0.05 = 1.86 / (Tf * M)
Now, solve for M:
M = 1.86 / (0.05 * Tf)
Substituting Tf = 271.60 K into the equation:
M = 1.86 / (0.05 * 271.60)
Let's calculate this value:
M = 1.86 / 13.58
M ≈ 0.137
The molar mass of boric acid is approximately 0.137 grams per mole.
To prepare 500 mL of an isotonic solution, we need to calculate the amount of boric acid required using the formula:
amount of boric acid (in grams) = volume of solution (in liters) x molality x molar mass
First, convert the volume of the solution to liters:
500 mL = 500 / 1000 = 0.5 liters
Now, calculate the amount of boric acid:
amount of boric acid = 0.5 * 0.05 * 0.137
amount of boric acid ≈ 0.003425 grams
Therefore, approximately 0.003425 grams of boric acid should be used to prepare a 500 mL isotonic solution.