Answer:
Option C. 90.0 g glucose per 500 cm3 water
Step-by-step explanation:
1.00 M (mol/dm3) aqueous solution of glucose (C6H12O6) means that 1 mole of glucose is dissolved in 1dm³ (1000cm³) of water.
The concentration of a substance is defined as the mole of solute per unit volume of solvent (water) in dm³.
Option A is wrong since it state: 180g of C6H12O6 per 1000cm³ of solution. Concentration is always expressed as per 1000cm³ or dm³ of water.
For option B:
We shall determine the number of mole of C6H12O6, followed by the concentration.
This is illustrated below:
Molar mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 180g/mol
Mass of C6H12O6 = 10g
Mole = Mass /Molar Mass
Mole of C6H12O6 = 10/180 = 0.056 mole.
Volume of water = 10cm³ = 10/1000 = 0.01dm³
Concentration = mole /Volume
Concentration = 0.056/0.01 = 5.6mol/dm³
Option B is wrong
For option C:
We shall determine the number of mole of C6H12O6, followed by the concentration.
This is illustrated below:
Molar mass of C6H12O6 = 180g/mol
Mass of C6H12O6 = 90g
Mole = Mass /Molar Mass
Mole of C6H12O6 = 90/180 = 0.5mole
Volume of water = 10cm³ = 500/1000 = 0.5dm³
Concentration = mole /Volume
Concentration = 0.5/0.5 = 1mol/dm³
Option C is correct.
For option D is wrong as well as it states 0.1g of C6H12O6 per cm³ of solution. Concentration is always per 1000cm³ or dm³ of water.
From the above illustrations, option C is the correct answer.
Explanation:
