Ncert Exercises & Solutions

If the concentration of glucose (C₆H₁₂O₆) in the blood is 0.9 g L^-1, then the molarity of glucose in the blood is:
1. 5 M
2. 50 M
3. 0.005 M
4. 0.05 M

Hint: Molarity unit is mol

L^{- 1}

Step 1:

In the given question, 0.9 g L^-1 means that 1000 mL (or 1L) solution contains 0.9 g of glucose.

The molar mass of glucose = 180 g/mol

Calculate the number of moles of glucose as follows:

number of moles = \(\frac{amount}{Molar \ mass}\)
= \(\frac{0.9}{180}\)
= 5×10^-3 mol glucose

Step 2:

Calculate the molarity of the solution as follows:

Molarity = \(\frac{number \ of \ moles \ of \ glucose}{Volume \ of \ the \ solution(L)}\)
= \(\frac{0.005}{1}\)
= 0.005 M
Hence, 1L solution contains 0.005 mole glucose or the molarity of glucose is 0.005 M.
Thus, option third is the correct answer.

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