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Question: A boy drinks 500 mL of \(9%\) glucose solution. The number of glucose molecules he has consumed is: ...

A boy drinks 500 mL of 99% glucose solution. The number of glucose molecules he has consumed is:
(A)- 0.5×10230.5\times {{10}^{23}}
(B)- 1.0×10231.0\times {{10}^{23}}
(C)- 1.5×10231.5\times {{10}^{23}}
(D)- 2.0×10232.0\times {{10}^{23}}

Explanation

Solution

The number of molecules consumed can be obtained, having known the concentration of the solution in percentage form (that is, weight by volume percent concentration). From which we can calculate the moles present in the solution, which constitutes the number of molecules along with the Avogadro’s number.

Complete step by step answer:
The concentration of the solution in terms of molality, molarity and normality as we already know, determine the amount of solute present in the solution. Similarly, we also have the percentage concentration used to express it.
The percentage concentration of the glucose solution is given to be 99%, that is, the weight by volume percentage concentration of the solution. So, we have 9 grams of glucose in 100 ml of solution.
Then, the mass of glucose in 500 ml of solution will be =9100×500=45g=\dfrac{9}{100}\times 500=45g
The molecular mass of a glucose molecule, C6H12O6=(6×12+12×1+6×16)=180g/mol{{C}_{6}}{{H}_{12}}{{O}_{6}}=(6\times 12+12\times 1+6\times 16)=180g/mol

Now, we can find the moles of glucose consumed =mass dissolvedmolecular mass=45180=0.25moles\text{=}\dfrac{\text{mass}\,\text{dissolved}}{\text{molecular}\,\text{mass}}\text{=}\dfrac{\text{45}}{180}=0.25moles
As the number of molecules in one mole =6.022 !!×!! 1023 molecules/mole\text{=6}\text{.022 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ molecules/mole}
Then, the number of molecules present in 0.25 moles =0.25×6.022×1023=1.5×1023=0.25\times 6.022\times {{10}^{23}}=1.5\times {{10}^{23}} molecules
Therefore, the number of glucose molecules the boy has consumed in 500ml of glucose solution is option (C)- 1.5×10231.5\times {{10}^{23}} molecules in 0.25 moles.
So, the correct answer is “Option C”.

Note: In the percentage form, the concentration be expressed in the following ways:
The weight by volume percentage, we have the mass of solute in 100 ml of solution.
The mass percentage (w/w)(w/w) is the percent of mass of solute to the total mass of solution or the mass (in grams) of solute in 100 grams of solution. And we have the volume percentage, with volume (in mL) of solute in 100 ml of solution.