Q.A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL$^{-1}$, then what shall be the molarity of the solution?

Concept: Molality Calculation — Molality depends only on the mass of solvent, not on volume or density.

Step 1: Interpret 10% w/w glucose

10 g glucose in 100 g solution → mass of water = 90 g = 0.090 kg.

Molar mass of glucose ($C_6H_{12}O_6$) = 180 g mol$^{-1}$.

Moles of glucose = $\frac{10}{180} = 0.0556$ mol.

Step 2: Molality

$m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{0.0556}{0.090} = 0.617$ mol kg$^{-1}$.

Step 3: Mole fractions

Moles of water = $\frac{90}{18} = 5.00$ mol.

Total moles = $0.0556 + 5.00 = 5.0556$ mol.

$x_{\text{glucose}} = \frac{0.0556}{5.0556} = 0.0110$

$x_{\text{water}} = 1 - 0.0110 = 0.9890$.

Step 4: Molarity from density …

Let’s unpack this step by step. The problem gives a “10% w/w” glucose solution — that means 10 grams of glucose are present in every 100 grams of the solution. The rest (90 g) is water, the solvent. This is the starting point for all three quantities.

1. Molality — moles of solute per kg of solvent

Molality ($m$) is defined as:

First, find moles of glucose. Glucose is $\text{C}_6\text{H}_{12}\text{O}_6$, molar mass = $6 \times 12 + 12 \times 1 + 6 \times 16 = 72 + 12 + 96 = 180 \text{ g mol}^{-1}$.

In 100 g of solution, we have 10 g glucose. So:

Mass of solvent (water) = $100 - 10 = 90 \text{ g} = 0.090 \text{ kg}$.

Thus:

2. Mole fraction of each component

Mole fraction ($x$) is moles of one component divided by total moles in the solution.

We already have moles of glucose = $0.05556$.

Moles of water: mass of water = 90 g, molar mass = 18 g mol$^{-1}$.

Total moles = $0.05556 + 5.00 = 5.05556$ mol.

Mole fraction of glucose:

Mole fraction of water:

Check: $0.0110 + 0.9890 = 1.0000$ — good. …

Method: Mass-Based Composition Conversion (w/w % → Molality → Mole Fraction → Molarity)

This method uses the mass percentage as the starting point, converting step-by-step using definitions of molality, mole fraction, and molarity.

Step 1: Interpret the 10% w/w label

• 10% w/w means 10 g of glucose in 100 g of solution.
• So, mass of glucose = 10 g
• Mass of water (solvent) = 100 g − 10 g = 90 g

Step 2: Calculate molality ($m$)

Formula:

• Molar mass of glucose ($C_6H_{12}O_6$) = $6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \text{ g/mol}$
• Moles of glucose = $\frac{10}{180} = 0.0556 \text{ mol}$
• Mass of solvent = $90 \text{ g} = 0.090 \text{ kg}$

Answer: Molality = 0.617 m

Step 3: Calculate mole fraction of each component

Formula:

• Moles of water = $\frac{90}{18} = 5.00 \text{ mol}$ (molar mass of water = 18 g/mol)
• Total moles = $0.0556 + 5.00 = 5.0556 \text{ mol}$

Answer:

• Mole fraction of glucose = 0.0110
• Mole fraction of water = 0.9890

Step 4: Calculate molarity ($M$)

Formula: …

🧠 Common Mistake #1: Confusing % w/w with % w/v or % v/v

The error:

Students treat “10% w/w” as 10 g of glucose in 100 mL of solution (which is % w/v) or 10 g in 100 g of solution (correct for w/w) but then incorrectly use volume for molality.

How to avoid:

• % w/w = mass of solute per 100 g of solution (not per 100 mL).
• Always write:

  10% w/w → 10 g glucose + 90 g water (total 100 g solution).

• Molality uses mass of solvent in kg, not mass of solution.

🧠 Common Mistake #2: Using density at the wrong step

The error:

Students plug density into molality calculation. Molality does not involve volume or density — only masses.

How to avoid:

• Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$
• Density is needed only for molarity (which uses volume of solution).
• Keep them separate:
  • Molality → use masses.
  • Molarity → use density to get volume from mass of solution.

🧠 Common Mistake #3: Forgetting to convert grams to kg for molality

The error:

Using mass of solvent in grams directly in the molality formula.

How to avoid:

• Always convert:

  $90 \, \text{g} = 0.090 \, \text{kg}$

• Formula:

🧠 Common Mistake #4: Wrong mole fraction formula

The error:

Using mass instead of moles, or forgetting that mole fraction of all components must sum to 1.

How to avoid:

• Mole fraction of glucose:

• Mole fraction of water:

• Always verify: $x_{\text{glucose}} + x_{\text{water}} = 1$

🧠 Common Mistake #5: Using density of pure water instead of solution density for molarity

The error:

Assuming density of solution = 1 g/mL (like pure water) when it’s given as 1.2 g/mL.

How to avoid:

• Read the problem carefully — density is of the solution, not solvent.
• For molarity:
  • Mass of 100 g solution → Volume = $\frac{100 \, \text{g}}{1.2 \, \text{g/mL}} = 83.33 \, \text{mL}$
  • Convert to L: $0.08333 \, \text{L}$ …