If the concentration of glucose  (C6H12O6)  in the blood is 0.9 g L–1, then the molarity of glucose in the blood is: 
1. 5 M
2. 50 M
3. 0.005 M
4. 0.05 M

Hint: Molarity unit is mol L-1
Step 1: 
In the given question, 0.9 g L-1 means that 1000 mL (or 1L) solution contains 0.9 g of glucose.

The molar mass of glucose  = 180 g/mol
Calculate the number of moles of glucose as follows: 

number of moles = \(\frac{amount}{Molar \ mass}\)
\(\frac{0.9}{180}\)
= 5×10-3 mol glucose 

Step 2: 

Calculate the molarity of the solution as follows: 

Molarity = \(\frac{number \ of \ moles \ of \ glucose}{Volume \ of \ the \ solution(L)}\) 
\(\frac{0.005}{1}\)
= 0.005 M
Hence, 1L solution contains 0.005 mole glucose or the molarity of glucose is 0.005 M.
Thus, option third is the correct answer.