Question:
If the concentration of glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)$ in blood is $0.72 \mathrm{~g} \mathrm{~L}^{-1}$, the molarity of glucose in blood is __________ $\times 10^{-3} \mathrm{M}$. (Nearest integer)
[Given: Atomic mass of $\mathrm{C}=12, \mathrm{H}=1, \mathrm{O}=16 \mathrm{u}$ ]
Solution:
$[$ Glu cose $]=\frac{\mathrm{C}(\mathrm{gm} / \ell)}{\mathrm{M}(\mathrm{gm} / \mathrm{mol})}=\frac{0.72}{180}=4 \times 10^{-3} \mathrm{M}$
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