Question

If the concentration of glucose (C₆H₁₂O₆) in blood is 0.9g L⁻¹, what will be the molarity of glucose in blood ?

A. 5 M

B. 50 M

C. 0.005 M

D. 0.5 M

Answer 1

The molarity of glucose in blood with a concentration of 0.9g L-1 is calculated using the molecular weight of glucose (180.156 g/mol) and results in a molarity of approximately 0.005 M, which can be expressed as 0.5 mM.

Step-by-step explanation:

To calculate the molarity of glucose in blood when the concentration is given as 0.9g L-1, we first need to know the molecular weight of glucose (C6H12O6). The molecular weight is the sum of the atomic weights of its components and for glucose, it is approximately 180.156 g/mol. Using this molecular weight, we can convert the given concentration in grams per liter to molarity, which is moles per liter (mol/L).

Here is the calculation:

Molarity (M) = Mass of solute (g/L) / Molecular weight of solute (g/mol)

M = 0.9 g/L / 180.156 g/mol

M ≈ 0.00499 mol/L

So, the molarity of glucose in blood is approximately 0.005 M, which can be rounded to 0.5 mM because molarity is often expressed in millimoles per liter (mM) in the context of blood measurements.