If the concentration of glucose (C_{6}H_{12}O | vedclass

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(A) The molar mass of glucose $(C_{6}H_{12}O_{6})$ is calculated as: $6 	imes 12 + 12 	imes 1 + 6 	imes 16 = 72 + 12 + 96 = 180 \ g \ mol^{-1}$.Mol

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$4$

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(A) The molar mass of glucose $(C_{6}H_{12}O_{6})$ is calculated as: $6 	imes 12 + 12 	imes 1 + 6 	imes 16 = 72 + 12 + 96 = 180 \ g \ mol^{-1}$.Molarity $(M)$ is defined as the number of moles of solute per liter of solution.$M = \frac{	ext{mass of solute (g)}}{	ext{molar mass (g/mol)} 	imes 	ext{volume of solution (L)}}$.Given concentration $= 0.72 \ g \ L^{-1}$,so for $1 \ L$ of solution,mass $= 0.72 \ g$.$M = \frac{0.72 \ g}{180 \ g \ mol^{-1} 	imes 1 \ L} = 0.004 \ mol \ L^{-1}$.$0.004 \ mol \ L^{-1} = 4 	imes 10^{-3} \ M$.

If the concentration of glucose $(C_{6}H_{12}O_{6})$ in blood is $0.72 \ g \ L^{-1}$,the molarity of glucose in blood is $..... \times 10^{-3} \ M$. (Nearest integer)
(Given: Atomic mass of $C=12, H=1, O=16 \ u$)

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If the concentration of glucose $(C_{6}H_{12}O_{6})$ in blood is $0.72 \ g \ L^{-1}$,the molarity of glucose in blood is $..... \times 10^{-3} \ M$. (Nearest integer)(Given: Atomic mass of $C=12, H=1, O=16 \ u$)