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Question: If the concentration of glucose ($C_6H_{12}O_6$) in blood is 0.72 $gL^{-1}$, the molarity of glucose...

If the concentration of glucose (C6H12O6C_6H_{12}O_6) in blood is 0.72 gL−1gL^{-1}, the molarity of glucose in blood is ____×10−3\_\_\_\_ \times 10^{-3}M (Nearest integer) [Given: Atomic mass of C = 12, H = 1, O = 16 u]

Answer

4

Explanation

Solution

Molar mass of glucose (C6H12O6C_6H_{12}O_6) is calculated as: (6×12)+(12×1)+(6×16)=72+12+96=180 g/mol(6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180 \text{ g/mol} Molarity is calculated using the formula: Molarity=Concentration (g/L)Molar Mass (g/mol)\text{Molarity} = \frac{\text{Concentration (g/L)}}{\text{Molar Mass (g/mol)}} Substituting the given values: Molarity=0.72 g/L180 g/mol=0.004 mol/L\text{Molarity} = \frac{0.72 \text{ g/L}}{180 \text{ g/mol}} = 0.004 \text{ mol/L} To express this in the format ____×10−3\_\_\_\_ \times 10^{-3}M: 0.004 M=4×10−3 M0.004 \text{ M} = 4 \times 10^{-3} \text{ M} The value to be filled in the blank is 4. The question asks for the nearest integer, which is 4.