Home / IB DP Chemistry -Structure 1.1 Introduction to the particulate nature of matters – IB Style Questions For HL Paper 1A

IB DP Chemistry -Structure 1.1 Introduction to the particulate nature of matters - IB Style Questions For HL Paper 1A -FA 2025

IBDP Chemistry HL Paper 1A & 1B - All Chapters

Question

The thin-layer chromatogram below was obtained from a mixture of naphthalene, C10H8, and naphthol, C10H8O. A polar silica surface was used as the stationary phase, and non-polar hexane served as the mobile phase.

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
What are the position and retardation factor (Rf) of naphthalene?
(A) y and 0.50
(B) y and 0.58
(C) x and 0.80
(D) x and 0.83
▶️ Answer/Explanation
Detailed solution

Chromatography principles:

  • Polar stationary phase (silica)
  • Non-polar mobile phase (hexane)
  • Less polar compounds travel further
  • Naphthalene (non-polar) travels further than naphthol (polar due to -OH group)

Naphthalene will be at position x (further from origin) with Rf = distance traveled / solvent front.

If x = 5.6 cm and solvent front = 7 cm, Rf = 5.6/7 = 0.80

✅ Answer: (C) – x and 0.80

Question

In the following unbalanced equation, X represents an element. Which oxide reacts with water as shown?
\( \dots + H_2O \rightarrow X(OH)_2 \)
A. \(Na_2O\)
B. \(MgO\)
C. \(NO_2\)
D. \(SO_3\)
▶️ Answer/Explanation
Detailed solution

The product \(X(OH)_2\) is a metal hydroxide with a 2+ metal cation (a typical group 2 metal). Such hydroxides are formed when a basic oxide of a group 2 metal reacts with water:
\( MO(s) + H_2O(l) \rightarrow M(OH)_2(aq) \)

For magnesium:
\( MgO(s) + H_2O(l) \rightarrow Mg(OH)_2(aq) \)

Check the options:
\(Na_2O\) is a group 1 oxide and forms \(NaOH\), not \(X(OH)_2\).
\(NO_2\) and \(SO_3\) are acidic oxides and react with water to form acids, not metal hydroxides.

Therefore, the oxide that reacts with water to give \(X(OH)_2\) is \(MgO\).
✅ Answer: (B)

Question

What is the sum of the coefficients when the equation is balanced with the lowest whole number ratio?
\[ \_\ \text{Na}_2\text{S}_2\text{O}_3(aq) + \_\ \text{HCl}(aq) \;\longrightarrow\; \_\ \text{S}(s) + \_\ \text{SO}_2(g) + \_\ \text{NaCl}(aq) + \_\ \text{H}_2\text{O}(l) \]
A.  \(6\)
B.  \(7\)
C.  \(8\)
D.  \(9\)
▶️ Answer/Explanation
Detailed solution

Begin with sodium. In \(\text{Na}_2\text{S}_2\text{O}_3\), there are \(2\) sodium atoms. To balance sodium, place a coefficient of \(2\) in front of \(\text{NaCl}\).

\[ \text{Na}_2\text{S}_2\text{O}_3 + \text{HCl} \rightarrow \text{S} + \text{SO}_2 + 2\text{NaCl} + \text{H}_2\text{O} \]

Sulfur: The left side has \(2\) sulfur atoms; the products have \(1\) in \(\text{S}\) and \(1\) in \(\text{SO}_2\). Oxygen: The left has \(3\); the right has \(2\) (in \(\text{SO}_2\)) and \(1\) (in \(\text{H}_2\text{O}\)), so oxygen is balanced.

Hydrogen: The product side has \(2\) hydrogens (in \(\text{H}_2\text{O}\)), so place a coefficient of \(2\) in front of \(\text{HCl}\).

Final balanced equation: \[ 1\ \text{Na}_2\text{S}_2\text{O}_3 + 2\ \text{HCl} \rightarrow 1\ \text{S} + 1\ \text{SO}_2 + 2\ \text{NaCl} + 1\ \text{H}_2\text{O} \]

Sum of coefficients: \[ 1 + 2 + 1 + 1 + 2 + 1 = 8 \]

✅ Answer: (C)

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