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Indian Navy Pilot Solutions — practice questions

Explore 16 free Solutions questions in Indian Navy Pilot Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Solutions questions and exercises with answers

Showing questions 1–16 of 16.

The van't Hoff factor $i$ for a $0.01$ M aqueous solution of NaCl, assuming complete dissociation, is approximately:Read the answer and worked solution →Molarity M is defined as:Read the answer and worked solution →Two volatile liquids A (P°_A = 100) and B (P°_B = 200) form ideal solution. For x_A = 0.4, total vapor pressure:Read the answer and worked solution →The van't Hoff factor (i) for sodium chloride (NaCl) in water is approximately:Read the answer and worked solution →Complete the two concentration definitionsRead the answer and worked solution →Molarity or molality?Read the answer and worked solution →True or false: concentrationRead the answer and worked solution →Naming the measureRead the answer and worked solution →Complete the meaning of colligativeRead the answer and worked solution →True or false: colligative propertiesRead the answer and worked solution →Tap every statement true of molalityRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Flip and pair: measure with its denominatorRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →

Worked practice questions

The van't Hoff factor $i$ for a $0.01$ M aqueous solution of NaCl, assuming complete dissociation, is approximately:

  1. $1$
  2. $0.5$
  3. $2$
  4. $3$
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Correct answer: C. $2$

NaCl dissociates in water: $\mathrm{NaCl} \rightarrow \mathrm{Na^+} + \mathrm{Cl^-}$. One formula unit produces two particles in solution. Van't Hoff factor $i = \dfrac{\text{actual moles of particles}}{\text{moles of solute formula units}}$. For complete dissociation of NaCl, $i = 2$. In reality, ion pairing slightly reduces $i$ below the theoretical value, especially at higher concentrations. At dilute $0.01$ M the assumption of complete dissociation is reasonable, and the experimental $i$ is close to $1.9$. The question asks for the theoretical value assuming complete dissociation, so 2 is the cleanest answer. Compare with $\mathrm{CaCl_2}$: dissociates into 3 ions, so $i = 3$ (option D) — common trap.

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Molarity M is defined as:

  1. Moles of solute per kg of solvent
  2. Volume of solute per litre
  3. Mass of solute per litre
  4. Moles of solute per litre of SOLUTION
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Correct answer: D. Moles of solute per litre of SOLUTION

Molarity (M) = moles of solute / litres of solution. *Note: this is per litre of SOLUTION, not solvent (different from molality).*

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Two volatile liquids A (P°_A = 100) and B (P°_B = 200) form ideal solution. For x_A = 0.4, total vapor pressure:

  1. 150
  2. 100
  3. 160 (= 100 × 0.4 + 200 × 0.6)
  4. 300
Show answer and solution

Correct answer: C. 160 (= 100 × 0.4 + 200 × 0.6)

P_total = P°_A x_A + P°_B x_B = 100 × 0.4 + 200 × 0.6 = 40 + 120 = 160 mmHg. (Linear in composition for ideal mixtures.)

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