Which of the following represents a 1.00 M (M = mol/dm3) aqueous solution of glucose (C6H12O6)? A. 180 g glucose per 1000 cm3 solution

B. 10.0 g glucose per 10.0 cm3 water

C. 90.0 g glucose per 500 cm3 water

D. 0.100 g glucose per cm3 solution

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.

-8/2y-8=5/y+4-7y+8/y^2-16

we move all terms to the left:

-8/2y-8-(5/y+4-7y+8/y^2-16)=0

we can not solve this equation

3

step-by-step explanation: