Answer:
The concentration of glucose 95.4 mg/dL:
Explanation:
[tex]Molarity = 1000 \times \text{milliMolarity}[/tex]
Concentration of glucose in blood = 5.3 milli Molar
This means that 5.3 milli moles are present in 1 l of solution.
5.3 milli moles = 0.0053 moles
So, from this we can say that 0.0053 moles of glucose are present in 1 L solution.
Mass of 0.0053 moles of glucose :
= 0.0053 mol × 180 g/mol = 0.954 g
(1 g = 1000 mg)
0.954 g = 954 mg
945 milli grams in 1 liter of the solution.
(1 L = 10 dL)
The concentration of glucose in mg/dL:
[tex]\frac{945 mg}{10 dL}=94.5 mg/dL[/tex]
