Question

Calculate the molarity of a solution made by dissolving 5.00 g of glucose (C6H12O6) in sufficient water to form exactly 100 mL of solution.

Answer 1

I think it’s 2.7 moldm-3

Answer 2

`0.28 mol {dm}^( - 3)`

Step-by-step explanation:

given; 5.00g equivalent to 100mL

5.00g = 100mL

convert 100mL to dm³

100mL is 100milliLitre

milli = 10^-3

100mL = 100 × (10^-3)

= 10^2 × 10^-3

=

`{10}^(2 - 3)`

= 10^-1 L

`1l = 1 {dm}^(3)`

therefore, 10^-1 L = 10^-1 dm³

Relative Atomic Mass of;

C=12, H=1, O=16

Molar mass of glucose (C6H12O6)

= 12×6 + 1×12 + 16×6

= 180g/mol

Now convert the mass given to mole using;

`mole \: = (mass)/(molar \: mass)`

`mole \: = (5.00)/(180)`

mole = 0.028 mol

therefore, 0.028 mol is equivalent to 10^-1 dm³

10^-1 dm³ = 0.028mol

divide both sides by 10^-1 to get 1dm³

`1 {dm}^(3) = \frac{0.028 \: mol}{ {10}^( - 1) }`

= 0.28 mol

`molarity \: = 0.28mol {dm}^( - 3)`