Q.Calculate the mole fraction of ethylene glycol ($C_2H_6O_2$) in a solution containing 20% of $C_2H_6O_2$ by mass.

Concept & Intuition

Mass percent tells you how many grams of solute are present in every 100 grams of solution. It’s a simple ratio:

The trick here is that the solvent (water) is given in moles, not grams. So the first step is always to convert moles of water to grams using its molar mass (18 g/mol). Once everything is in grams, the calculation is straightforward.

Step-by-step solution

1. Find the mass of water (solvent) We have 2.0 moles of water. The molar mass of water is $18.0\ \text{g/mol}$.

2. Find the total mass of the solution The solution contains the solute (4.0 g of X) plus the solvent (36 g of water).

3. Calculate the mass percent

Step 1 — Write each concentration measure with its units.

1. Mole fraction

Moles divided by moles $\Rightarrow$ unitless (and it always lies between 0 and 1).

2. Mass percent (W/W)

Grams divided by grams $\Rightarrow$ unitless (the "%" is a pure number, not a unit).

3. Molality

Moles divided by mass — two different kinds of quantity $\Rightarrow$ has units.

4. Molarity

Moles divided by volume $\Rightarrow$ has units.

Step 2 — Apply the test.

The unitless pair is therefore mole fraction and mass percent (W/W).

Step 3 — Eliminate.

• (B) Molality has units ($\mathrm{mol\,kg^{-1}}$) — fails.
• (C) Molality has units — fails.
• (D) Both molality and molarity have units — fails outright.

Only (A) lists two genuinely dimensionless measures.

Aside worth knowing: the fact that mole fraction, molality and mass percent are all temperature-independent (they involve no volume, which expands on heating) — whereas molarity is temperature-dependent — is a closely related and frequently examined point. But the specific question here is about units, and on that test molality is excluded while mass percent is included.

ANSWER: A

Concept & Intuition

We have a haloalkane (an alkane with one chlorine atom replacing a hydrogen). The problem tells us that chlorine makes up 55% of the mass of one molecule. That means if we know the mass of one mole of the compound, we can find how many moles of Cl are in a sample. Since each molecule has exactly one Cl atom, the number of Cl atoms equals the number of molecules. So the plan: find the molar mass from the percentage, then convert 0.1 g to moles, then to atoms via Avogadro’s number.

Step-by-step solution

1. Relate percentage to molar mass Let $M$ be the molar mass of the haloalkane (in g/mol). One mole of the compound contains one mole of chlorine atoms, which has mass $35.5$ g. The percentage by mass of chlorine is:

So:

Solving:

(We can keep it as $\frac{35.5}{0.55}$ for now.)

2. Find moles of haloalkane in 0.1 g Moles of compound:

Simplify:

3. Number of molecules (and thus Cl atoms) Since each molecule has one Cl atom, the number of Cl atoms is:

Compute step by step:

Multiply by Avogadro’s number:

4. Match with options The value $9.328 \times 10^{20}$ matches option (C) exactly (option D is off by a factor of 1000, a common mistake if you forget to convert grams to moles properly).

Step 1 — The concept.

Mixtures are classified by whether their composition is the same at every point:

• Homogeneous mixture (solution): solute particles are of molecular/ionic size ($< 1\,$nm), uniformly dispersed. Every sample drawn from anywhere has the same composition. Only one phase is visible.
• Heterogeneous mixture: two or more distinguishable phases; composition varies from point to point.

Step 2 — Test each option.

(A) Sand and water — sand particles are large and insoluble; they settle to the bottom. Two visible phases → heterogeneous ✗

(B) Grains and pulses with stone — plainly separable solids, each retaining its identity; a scoop from one corner differs from another → heterogeneous ✗

(C) Mixture of oil and water — oil is non-polar, water is polar; they are immiscible and separate into two layers with a clear meniscus → heterogeneous ✗

(D) Dilute aqueous solution of sugar — sucrose is polar and hydrogen-bonds with water, so it dissolves completely into individual molecules dispersed at random. Every drop tastes equally sweet and has the same concentration → homogeneous ✓

Step 3 — Commit. Only (D) has uniform composition throughout.

ANSWER: D

The concept here is mole fraction — a way to express concentration in terms of the number of particles (moles) rather than mass. In a mixture, the mole fraction of a component is simply the number of moles of that component divided by the total number of moles of all components. It’s dimensionless and always lies between 0 and 1.

Why does this matter? Because mole fraction directly relates to partial pressures in gases and colligative properties in solutions. For this problem, we just need to convert the given masses into moles using molar masses, then compute the ratio.

1. Find the moles of water. Water ($H_2O$) has a molar mass of $18 \, \text{g/mol}$. Given $18 \, \text{g}$ of water:

2. Find the moles of ethyl alcohol. Ethyl alcohol ($C_2H_5OH$) has a molar mass of $46 \, \text{g/mol}$ (carbon: $2 \times 12 = 24$, hydrogen: $6 \times 1 = 6$, oxygen: $16$, total $24+6+16=46$). Given $414 \, \text{g}$ of alcohol:

3. Calculate total moles.

4. Compute mole fraction of water.

Step 1 — Set up the bcc geometry.

A bcc unit cell has:

• $8$ corner atoms, each shared by $8$ cells $\Rightarrow 8\times\tfrac18 = 1$ atom
• $1$ atom fully inside at the body centre $\Rightarrow 1$ atom

Step 2 — Relate radius to edge length.

In bcc the atoms touch along the body diagonal, not along the edge. The body diagonal of a cube of edge $a$ has length $\sqrt3\,a$, and it contains $4$ radii (corner atom radius + full central atom + corner atom radius):

Step 3 — Compute the packing efficiency.

So the atoms fill $68\%$ of the cell.

Step 4 — Vacant space.

(For comparison: fcc/ccp is $74\%$ filled → $26\%$ vacant; simple cubic is $52.4\%$ filled → $47.6\%$ vacant. Option (B) $46\%$ is the trap meant for simple cubic, and (A) $23\%$ echoes fcc.)

ANSWER: C