A solution is prepared by dissolving 18g of water in 150g of glucose,the resulting solution was found to have boiling point of 100.34°c. Calculate the molar mass of glucose.
I know the boiling point constant for H2O (0.512 degreed C/molal) but not for glucose; therefore, my assumption is that you are dissolving the glucose in the water and not the other way. In addition, I do not believe your numbers are right. I think you meant that you were dissolving 18 g glucose in 150 g water.
Please correct your problem. You SHOULD proof before you publish.
Very good
To calculate the molar mass of glucose, we first need to calculate the total number of moles of solute (glucose) dissolved in the solution.
Step 1: Calculate the number of moles of water (H2O)
Molar mass of water (H2O) = 2(1.008 g/mol) + 16.00 g/mol = 18.02 g/mol
Number of moles of water (H2O) = mass of water (g) / molar mass of water (g/mol)
Number of moles of water (H2O) = 18 g / 18.02 g/mol ≈ 0.999 moles
Step 2: Calculate the number of moles of solute (glucose)
Number of moles of solute (glucose) = mass of solute (g) / molar mass of solute (g/mol)
Since we dissolve 18g of water in 150g of glucose, the mass of solute (glucose) is equal to the mass of glucose in the solution, which is 150g.
Number of moles of solute (glucose) = 150 g / molar mass of glucose (g/mol)
Step 3: Calculate the number of moles of the resulting solution
In the resulting solution, the total number of moles of water and glucose together is given by the sum of the moles of water and moles of glucose.
Total number of moles = number of moles of water + number of moles of glucose
Since the boiling point of the resulting solution is given as 100.34°C, which is equal to the boiling point of pure water (100°C) at 1 atmosphere of pressure, this implies that the solution is non-volatile and follow Raoult's law. According to Raoult's law, the boiling point elevation of the solution is directly proportional to the molality of the solute.
Molality (m) = moles of solute / mass of solvent (kg)
Since we are given the mass units in grams, we need to convert the mass of water to kg.
Mass of water (kg) = mass of water (g) / 1000
Plugging in the values, we have:
100.34°C - 100.00°C = K_b * m
Where K_b is the molal boiling point elevation constant for water, which is equal to 0.512 °C/m.
0.34°C = (0.512 °C/m) * (number of moles of solute / (mass of water (g) / 1000))
Now we can solve for the number of moles of solute (glucose):
Number of moles of solute (glucose) = (0.34°C * (mass of water (g) / 1000)) / (0.512 °C/m)
Number of moles of solute (glucose) = (0.34°C * 18 g / 1000) / (0.512 °C/m)
Number of moles of solute (glucose) ≈ 0.011 moles
Step 4: Calculate the molar mass of glucose
Molar mass of glucose (g/mol) = mass of solute (g) / number of moles of solute (glucose)
Molar mass of glucose = 150 g / 0.011 moles ≈ 13,636 g/mol
Therefore, the molar mass of glucose is approximately 13,636 g/mol.
To calculate the molar mass of glucose (C6H12O6), we need to use the colligative property known as boiling point elevation. The formula used to calculate the boiling point elevation is:
ΔTb = Kbm,
where:
ΔTb is the boiling point elevation,
Kb is the boiling point elevation constant (which is different for each solvent),
m is the molality of the solution (moles of solute per kilogram of solvent).
First, we need to calculate the molality (m) of the solution. Molality is calculated by dividing the moles of solute by the mass of the solvent (in kg).
Given:
Mass of glucose (solute) = 150 g
Mass of water (solvent) = 18 g
To convert the mass of water to kg:
Mass of water (in kg) = 18 g / 1000 = 0.018 kg
Now, we can calculate the molality (m):
m = moles of solute / mass of solvent (in kg)
To calculate the moles of solute, we need to know how many moles of glucose are present in 150 g.
Mass of glucose (solute) = 150 g
Molar mass of glucose (C6H12O6) = ?
The molar mass of glucose can be calculated using the following formula:
Molar mass (g/mol) = mass (g) / moles.
Rearranging the formula, we have:
Moles (mol) = mass (g) / molar mass (g/mol)
Since the mass of glucose is given as 150 g, we can substitute this value into the equation and solve for moles:
Moles = 150 g / molar mass
Now, we have moles of solute (glucose) and the mass of the solvent (water) in kg. We can calculate the molality (m):
m = moles of solute / mass of solvent (in kg)
Finally, we can calculate the boiling point elevation (ΔTb) using the boiling point elevation constant (Kb) for water, which is 0.52 °C/m. In this case:
ΔTb = Kb * m
The given boiling point is 100.34 °C, which is the elevated boiling point. The normal boiling point of pure water at sea level is 100 °C.
Boiling point elevation (ΔTb) = elevated boiling point - normal boiling point
Now we can solve for the molar mass of glucose using the equation:
ΔTb = Kb * m = Kbf * molality,
where Kbf is the boiling point elevation constant for the solvent (water).
Molar mass of glucose can be calculated as follows:
Molar mass of glucose (C6H12O6) = mass (g) / moles.
Substituting the values into the formula above, we can solve for the molar mass of glucose.