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

Answer:

$0.28 mol {dm}^{ - 3}$

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;

$mole \: = \frac{mass}{molar \: mass}$

$mole \: = \frac{5.00}{180}$

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

$molarity \: = 0.28mol {dm}^{ - 3}$