Concentration of glucose `(C_(6)H_(12)O_(6))` in normal blood is approximately 90 mg per 100 mL. What is the molarity of glucose in blood ?

Answer

0.005 M

Explanation

Mass of glucose = 90 mg `= (90)/(1000) = 0.09 g`
No. of moles of glucose `= ("Mass of glucose")/("Molar mass")=(("0.09 g"))/(("180 g mol"^(-1)))=0.005` mol
Volume of solution = 100 mL = 0.1 L
Molarity of solution (M) `= ("No. of moles of glucose")/("Volume of solution in litres")=(("0.005 mol"))/(("0.1 L"))="0.005 mol L"^(-1)=0.005M`