Answer:
[tex] 0.28 mol {dm}^{ - 3} [/tex]
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
=
[tex] {10}^{2 - 3} [/tex]
= 10^-1 L
[tex]1l = 1 {dm}^{3} [/tex]
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;
[tex]mole \: = \frac{mass}{molar \: mass} [/tex]
[tex]mole \: = \frac{5.00}{180} [/tex]
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
[tex]1 {dm}^{3} = \frac{0.028 \: mol}{ {10}^{ - 1} } [/tex]
= 0.28 mol
[tex]molarity \: = 0.28mol {dm}^{ - 3} [/tex]
