Colligative Properties NEET 2016: Official Solution

What was the NEET 2016 colligative properties question on vapour pressure?

The NEET 2016 colligative properties question supplied 6.5 g of an unknown solute dissolved in 100 g water. Solution vapour pressure equals 732 mm at exactly 100 °C, Kb is given as 0.52 K kg mol⁻¹, and the task is to select the new boiling point from integer choices clustered around 100–103 °C.

This method based on the official NEET 2016 key lets you solve any NEET colligative-properties numerical that gives vapour pressure at 100 °C and Kb by first finding M_solute exactly as 31.75 g mol⁻¹ then computing ΔTb = 1.06 °C to pick the correct integer boiling point.

What is the official step-by-step solution for the NEET 2016 colligative properties question?

At 100 °C, P⁰ for pure water is 760 mm. Relative lowering is therefore (760 – 732)/760 = 28/760. This equals the mole fraction of solute.

Apply the exact relation given in the official key:

\frac{760 - 732}{760} = \frac{6.5 \times 18}{M \times 100}

Solving yields M_solute = 31.75 g mol⁻¹.

Next convert to molality for the ebullioscopy step:

m = \frac{6.5 \times 1000}{31.75 \times 100} = 2.047 {mol kg}^{- 1}

Then

Δ T_{b} = 2.047 \times 0.52 = 1.064^{\circ} C \approx 1.06^{\circ} C

Boiling point of solution = 100 + 1.06 = 101.06 °C. The nearest option is 101 °C.

What common mistake produces one of the wrong options in the NEET 2016 colligative properties question?

Treating the given mass of solute as if it directly gives molality without first solving for M from the vapour-pressure equation leads to wrongly inserting 6.5 g and 100 g into the Kb formula as though molar mass were 18 or 1. This produces ΔTb ≈ 2 °C or 3 °C and lands on 102 °C or 103 °C.

The correct sequence must derive M = 31.75 before any molality step. Skipping it breaks the link between the two colligative effects.

How are relative vapour pressure lowering and boiling point elevation linked in the NEET 2016 question?

Relative lowering equals n_solute / (n_solute + n_solvent) which simplifies for dilute aqueous solutions to (w/M) × (18/W). The official solution sets (760 – 732)/760 equal to (6.5 × 18)/(M × 100) and solves for M first.

Molality then re-uses that identical M:

m = \frac{w_{solute} \times 1000}{M_{solute} \times w_{solvent}}

The same numerical value of M therefore appears in both halves of the problem.

Using M = 31.75 g mol⁻¹ gives m ≈ 2.05 mol kg⁻¹ and ΔTb = 1.06 °C.
Using any other M yields ΔTb off by 0.5–2 °C.
Final rounding of 1.06 °C selects 101 °C among the integer choices.

This exact algebraic linkage is what the 2016 paper tested. Write both equations side-by-side on your answer sheet so the shared M is impossible to miss.

What are two similar colligative properties questions to practice next?

Question 1: 5 g of solute in 50 g water lowers vapour pressure by 2 % at 100 °C. Find molar mass, then ΔTb if Kb = 0.5 K kg mol⁻¹.

Relative lowering = 0.02 = (5 × 18)/(M × 50).

M = 90 g mol⁻¹.

m = (5 × 1000)/(90 × 50) = 1.11 mol kg⁻¹.

ΔTb = 1.11 × 0.5 = 0.555 °C, so boiling point = 100.555 °C. The steps mirror the 2016 method exactly.

Question 2: A solution boils at 100.52 °C with Kb = 0.52 K kg mol⁻¹. Find molar mass if 2.6 g solute is dissolved in 200 g water.

ΔTb = 0.52 °C, therefore m = ΔTb / Kb = 1 mol kg⁻¹.

1 = (2.6 × 1000)/(M × 200).

M = 13 g mol⁻¹.

Both problems again demand that you calculate M before feeding it into the second colligative equation.

Why is the NEET 2016 colligative properties question tagged hard and how do you finish it in 120 seconds?

The hard tag (tier 3) comes because it merges two colligative equations and demands accurate arithmetic on 31.75. Split the 120 seconds like this: 40 s for the M calculation from relative lowering, 40 s for molality and ΔTb, 40 s cross-check against options.

Never round M before final multiplication. Keep 31.75 × 100 exact until the last step.

What checklist should you use before the next mock on colligative properties?

P⁰ at 100 °C is always 760 mm for water.
M_solute must be calculated from given VP before any molality.
ΔTb result 1.06 °C maps only to 101 °C option.
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Frequently asked questions

What was the exact colligative properties NEET 2016 question?

It involved 6.5 g of unknown solute in 100 g of water with vapour pressure 732 mm at 100 °C and Kb = 0.52 K kg mol⁻¹. The task was to find the boiling point of the solution from given options.

How to find molar mass in colligative properties NEET 2016 problem?

Apply the relative lowering of vapour pressure formula (760-732)/760 = (6.5 × 18)/(M × 100). Solving this equation gives the molar mass of solute as 31.75 g mol⁻¹.

What is the boiling point in NEET 2016 colligative properties question?

After determining molality as approximately 2.05 mol kg⁻¹, multiply by Kb of 0.52 to get ΔTb = 1.06 °C. Thus the boiling point is 101.06 °C and the correct option is 101 °C.

Why did many students get the colligative properties NEET 2016 question wrong?

They skipped calculating the molar mass from the vapour pressure data and directly used the given masses in the boiling point elevation formula. This leads to incorrect ΔTb values of around 2 or 3 degrees instead of 1.06.

How are relative vapour pressure lowering and boiling point elevation linked in NEET 2016?

The same molar mass M derived from relative lowering (760-732)/760 = (6.5×18)/(M×100) is reused in molality m = (6.5×1000)/(M×100). This gives m≈2.05 and ΔTb=1.06°C, selecting 101°C.

Colligative Properties NEET 2016: Official Vapour Pressure Solution