If the concentration of glucose (C$$_6$$H$$_{12}$$O$$_6$$) in blood is 0.72 gL$$^{-1}$$, the molarity of glucose in blood is ___ $$\times 10^{-3}$$M (Nearest integer):
[Given : Atomic mass of C = 12, H = 1, O = 16 u]

The concentration of glucose is given as $$0.72 \text{ g L}^{-1}$$. The molar mass of glucose $$\text{C}_6\text{H}_{12}\text{O}_6$$ is $$6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 \text{ g/mol}$$.

Molarity is defined as moles of solute per litre of solution. So the molarity is $$\frac{0.72}{180} = 0.004 \text{ mol/L} = 4 \times 10^{-3} \text{ M}$$.

The answer is $$4$$.