Answer:
Here's what I get.
Explanation:
At the end of the reaction you will have a solution of the alcohol in THF.
The microdistillation procedure will vary, depending on the specific apparatus you are using, but here is a typical procedure.
- Transfer the solution to a conical vial.
- Add a boiling stone.
- Attach a Hickman head (shown below) and condenser.
- Place the assembly in in the appropriate hole of an aluminium block on top of a hotplate stirrer.
- Begin stirring and heating at a low level so the THF (bp 63 °C) can distill slowly.
- Use a Pasteur pipet to withdraw the THF as needed.
- When all the THF has been removed, raise the temperature of the Al block and distill the alcohol (bp 143 °C).
Here is the complete question.
Glycerol (C3H8O3), also called glycerine, is widely used in the food and pharmaceutical industries. Glycerol is polar and dissolves readily in water and polar organic solvents like ethanol. Calculate the mole fraction of the solvent in a solution that contains 1.61 g glycerol dissolved in 22.60 mL ethanol (CH3CH2OH; density = 0.7893 g/mol). Round to four significant digits
Answer:
0.9567 mol
Explanation:
Given that:
mass of glycerol = 1.61 g
molar mass of glycerol = 92.1 g/mol
no of mole = 
∴ number of moles of glycerol (
) = 
= 0.0175 mol
Volume of ethanol = 22.60 mL
Density of ethanol = 0.7893 g/mL
Since Density = 
∴ mass of ethanol = density of ethanol × volume of ethanol
mass of ethanol = 0.7893 g/mL × 22.60 mL
mass of ethanol = 17.838 g
Number of moles of ethanol 
= 
= 0.387 mole
∴ the mole fraction of the solvent can be determined as:
= 0.95673671199
≅ 0.9567 mol
∴ The mole fraction of the solvent in a solution that contains 1.61 g glycerol dissolved in 22.60 mL ethanol is = 0.9567 mol
The inverted cone, hope it helps .
Answer:
HC₂H₃O₂ + NaHCO₃ —> NaC₂H₃O₂ + CO₂ + H₂O
The coefficients are: 1, 1, 1, 1, 1
Explanation:
_HC₂H₃O₂ + _NaHCO₃ —> _NaC₂H₃O₂ + _CO₂ + _H₂O
To balance an equation, we simply do a head count of the individual elements and ensure they are balanced on both side.
For the above equation, we shall balance it as :
HC₂H₃O₂ + NaHCO₃ —> NaC₂H₃O₂ + CO₂ + H₂O
Reactant:
H = 5
C = 3
O = 5
Na = 1
Product:
H = 5
C = 3
O = 5
Na = 1
From the above, we can see that each element is the same on both side of the equation. Thus the equation is already balanced
HC₂H₃O₂ + NaHCO₃ —> NaC₂H₃O₂ + CO₂ + H₂O
The coefficients are: 1, 1, 1, 1, 1
