Q.If the concentration of glucose ($C_6H_{12}O_6$) in blood is $0.9\ \text{g L}^{-1}$, what will be the molarity of glucose in blood?

The key idea is molarity calculation: molarity (M) = moles of solute per litre of solution.

Step 1: Find the molar mass of glucose ($C_6H_{12}O_6$).

$6 \times 12 + 12 \times 1 + 6 \times 16 = 72 + 12 + 96 = 180\ \text{g mol}^{-1}$.

Step 2: Convert the given concentration to moles per litre.

Moles of glucose in 1 L = $\frac{0.9\ \text{g}}{180\ \text{g mol}^{-1}} = 0.005\ \text{mol}$. …

The question asks for molarity, not molality — a common point of confusion. Molarity ($M$) is defined as the number of moles of solute per litre of solution. Here, the concentration is already given in grams per litre ($\text{g L}^{-1}$), which is a mass concentration. Converting that to molarity is straightforward: divide by the molar mass of glucose.

Glucose has the formula $C_6H_{12}O_6$. Let’s calculate its molar mass quickly:

• Carbon: $6 \times 12 = 72$
• Hydrogen: $12 \times 1 = 12$
• Oxygen: $6 \times 16 = 96$
• Total: $72 + 12 + 96 = 180\ \text{g mol}^{-1}$

Now, the given mass concentration is $0.9\ \text{g L}^{-1}$. That means every litre of blood contains $0.9$ grams of glucose.

1. Find moles of glucose per litre

   Moles = mass / molar mass = $0.9\ \text{g} / 180\ \text{g mol}^{-1} = 0.005\ \text{mol}$.

2. Molarity is moles per litre

   Since we already have the mass per litre, the moles per litre is exactly $0.005\ \text{mol L}^{-1}$, which is $0.005\ \text{M}$. …

Method: Molarity from Mass Concentration

Concept first: Molarity ($M$) tells us the number of moles of solute per litre of solution.

We are given mass per litre ($0.9\ \text{g L}^{-1}$). To convert to moles per litre, we need the molar mass of glucose.

Steps

Step 1 — Find molar mass of glucose ($C_6H_{12}O_6$)

• Carbon: $6 \times 12 = 72$
• Hydrogen: $12 \times 1 = 12$
• Oxygen: $6 \times 16 = 96$

Molar mass = $72 + 12 + 96 = 180\ \text{g mol}^{-1}$

Step 2 — Convert mass concentration to molarity

Molarity ($M$) = $\frac{\text{mass per litre (g/L)}}{\text{molar mass (g/mol)}}$ …

Step 1: Understand what is being asked

We are given:

• Solute: glucose, $C_6H_{12}O_6$
• Concentration: $0.9\ \text{g L}^{-1}$ (mass per volume)
• We need: Molarity (moles per litre)

Molarity $= \frac{\text{moles of solute}}{\text{volume of solution in litres}}$

So the first step is to convert grams per litre into moles per litre.

Step 2: Calculate molar mass of glucose

Glucose: $C_6H_{12}O_6$

• Carbon: $6 \times 12 = 72$
• Hydrogen: $12 \times 1 = 12$
• Oxygen: $6 \times 16 = 96$

Molar mass $= 72 + 12 + 96 = 180\ \text{g mol}^{-1}$

Step 3: Convert given concentration to molarity

Given: $0.9\ \text{g L}^{-1}$

Moles per litre $= \frac{0.9}{180} = 0.005\ \text{mol L}^{-1}$

So molarity $= 0.005\ \text{M}$

Correct answer: (iii) 0.005 M

Common Mistakes & How to Avoid Them

✗ Mistake 1: Confusing molality with molarity

• What students do: They try to use mass of solvent or density, which is not given.
• Why it’s wrong: Molarity uses volume of solution (given as 1 L here). Molality uses mass of solvent (in kg).
• How to avoid: Read the unit carefully — $\text{g L}^{-1}$ is a mass/volume concentration, directly convertible to molarity.

✗ Mistake 2: Forgetting to convert grams to moles

• What students do: They take $0.9\ \text{g L}^{-1}$ and think it’s already moles.
• Why it’s wrong: Molarity is moles per litre, not grams per litre.
• How to avoid: Always ask: “Is this in grams or moles?” If grams, divide by molar mass.

✗ Mistake 3: Incorrect molar mass of glucose

• What students do: Use $C_6H_{12}O_6$ as $6+12+6 = 24$ or forget to multiply atomic masses.
• Why it’s wrong: Each element’s atomic mass must be multiplied by its subscript.
• How to avoid: Write out the calculation step-by-step: $6 \times 12 = 72$, $12 \times 1 = 12$, $6 \times 16 = 96$, then sum.

✗ Mistake 4: Decimal error in division

• What students do: $0.9 \div 180 = 0.005$ but they write $0.05$ or $0.5$.
• Why it’s wrong: Misplacing the decimal changes the answer by a factor of 10 or 100. …