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Sugar cane may be used as a renewable source of ethanol, C2H5OH, via the conversion of sucrose, C12H22O11, according to the following unbalanced equation:
Hydrogen iodide gas, HI, forms when hydrogen, H2 and iodine, I2 react.
= + 26.5 kJ/mole
Unbalanced equation for the formation of ethanol
When two or more reactant with different chemical composition forms a product of different chemical composition then it would be termed as chemical reaction. From the above reaction, it is apparent that sucrose and water are mixed together to form ethanol and hence, it is a chemical reaction.
Grams of ethanol formed
Molar mass ethanol
From the reaction,
1 mole of sucrose produces = 4 moles of ethanol
1 mole of sucrose produces 4 moles of ethanol = 4 * 46 grams of ethanol
0.750 moles of sucrose would forms grams of ethanol = (0.750 * 4 * 46)/1 = 138 grams
Grams of ethanol formed
Molar mass sucrose
1 mole (342 grams) of sucrose would produce 4 moles of ethanol = 4*46 grams of ethanol
37.5 grams of sucrose would produce grams of ethanol = (4*46*37.5)/342 = 20.2 grams
Theoretical yield of ethanol in grams
1 mole (342 grams) of sucrose would produce 4 moles of ethanol = 4*46 grams of ethanol
57.5 grams of sucrose would produce grams of ethanol = (4*46*57.5)/342 = 30.93 grams
Therefore, theoretical yield of ethanol in grams is 30.93 grams.
Percentage of yield of ethanol for the reaction of 57.5 gram of sucrose
Percentage of yield of ethanol for the reaction
Volume ( )
Idea Gas law
Here,
Milligrams of 0.80% w/v nalorphine solution
Obtained dose = 2.5 mg
0.80 % w/v means,
0.80 grams of solute in 100 ml
2.5mg of solute = (100*2.5*10-3)/0.80 = 0.3125 ml
Dipole – dipole forces – Molecules with net dipole moment possess
Hydrogen bonding – Molecule contains -OH groups possess
From the structure of dimethyl ether and ethanol, the intermolecular forces can be seen as
Intermolecular forces in dimethyl ether (CH3OCH3) = Dipole- dipole forces
Intermolecular forces in ethanol (C2H5OH) = hydrogen bonding
Hydrogen bonding intermolecular force is higher than dipole- dipole force and hence, the boiling point of ethanol would be greater than boiling point of dimethyl ether.
Molarity of solution of nitric acid = 0.250 M
Concentration (commercially available)
Volume of nitric acid
Hence, in order to prepare the solution of 500 ml volume of the commercially available concentration of nitric acid, 7.8125 ml of 16M of nitric acid needs to be diluted up to a volume of 750ml.
Molarity of solution = 0.050 M
Molar mass of sucrose = 342.2965 per gram mole
Therefore, the mass of sucrose required 4.27 grams to prepare the solution.
Therefore, in order to prepare the solution of 250ml volume, 4.28 grams of sucrose needs to add in 250 ml of water to get the desired molarity of solution i.e. 0.05M.
Lowering of freezing point of a solvent is directly proportional to the number of solute molecules (moles or ions). Hence, the compound which has maximum number of solute ions would lower the freezing point of 2kg of water.
Dissolution of compounds
(Total ions after dissolution 2)
(Total ions after dissolution 3)
It is apparent that would provide 3 ions after dissolution and therefore, in order to lower the freezing point of 2 kg of water, the best suitable method is to use 0.20 moles of Ba(OH)2
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