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CHEMISTRY, M2 EQ-Bank 13

  1. If you have a solution with a concentration of 1.2 mol/L and need 0.6 moles of solute, what volume of solution is required?   (1 mark)
  1. Explain why knowing the volume of a solution is important when preparing specific concentrations in laboratory settings.   (2 marks)
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a.    \(0.5\ \text{L}\)

 

b.    The exact volumes of solutions must be known as:

→ The concentration of each solution must be precise, as it affects the molar ratios and yields.

→ Miscalculating volume would lead to incorrect concentrations, impacting experimental results and validity.

Show Worked Solution

a.    \(V=\dfrac{n}{c}=\dfrac{0.6}{1.2}=0.5\ \text{L}\)

 

b.    The exact volumes of solutions must be known as:

→ The concentration of each solution must be precise, as it affects the molar ratios and yields.

→ Miscalculating volume would lead to incorrect concentrations, impacting experimental results and validity.

CHEMISTRY, M2 EQ-Bank 5

The compound potassium nitrate has the formula \(\ce{KNO3}\).

  1. A student makes a solution of this compound by dissolving 40.0 g in 250.0 mL of distilled water. Calculate the concentration of this solution in mol L\(^{-1}\).   (2 marks)
  1. The student now requires 500.0 mL of a 5.0% (w/v) solution. What volume of the solution in part (a) is required to make this?   (3 marks)
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a.    \(1.58\ \text{mol L}^{-1}\)

b.    \(0.156\ \text{L}\)

Show Worked Solution

a.    \(n\ce{(KNO3)}= \dfrac{m}{MM} = \dfrac{40.0}{39.10 + 14.01 + 3(16.00)} =0.396\ \text{mol}\)

Concentration of solution \(=\dfrac{n}{V} = \dfrac{0.396}{0.25} = 1.58\ \text{mol L}^{-1}\)

 

b.    A 5% solution requires 5 grams of solute in 100 mL of solution.

Therefore a 500 mL solution requires 25 grams of solute.

\(n\ce{(KNO3)} = \dfrac{25}{101.11} = 0.247\ \text{mol}\)

\(c\ce{(KNO3)} = \dfrac{n}{V} = \dfrac{0.247}{0.5} = 0.494\ \text{mol L}^{-1}\)

\(c_1V_1\) \(=c_2V_2\)  
\(1.58 \times V_1\) \(=0.494 \times 0.5\)  
\(V_1\) \(=\dfrac{0.247}{1.58}\)  
\(V_1\) \(=0.156\ \text{L}\)  

CHEMISTRY, M2 EQ-Bank 2 MC

What volume of 0.50 M hydrochloric acid  is required to neutralise 25.0 mL of 0.40 M sodium hydroxide?

\(\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}\)

  1. \(15\ \text{mL}\)
  2. \(20\ \text{mL}\)
  3. \(25\ \text{mL}\)
  4. \(30\ \text{mL}\)
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\(B\)

Show Worked Solution

→ Calculate the number of moles of \(\ce{NaOH}\):

\(\ce{n(NaOH)=c \times V} = 0.4 \times 0.025 = 0.0100\ \text{mol}\)

→ Since the ratio of hydrochloric acid to sodium hydroxide is \(1:1\), the number of moles of hydrochloric acid required is also 0.0100 mol.

→ Calculate the volume of hydrochloric acid required:

\(\ce{V(HCl)= \dfrac{\text{n}}{\text{c}}}= \dfrac{0.0100}{0.5}=0.02\ \text{L}\ =20\ \text{mL}\)

\(\Rightarrow B\)

CHEMISTRY, M2 2013 HSC 17* MC

A 25.0 mL sample of a 0.100 mol L\(^{-1}\) hydrochloric acid solution completely reacted with 23.4 mL of sodium hydroxide solution.

\(\ce{HCl + NaOH -> NaCl + H2O}\)

\(\ce{CH3COOH + NaOH -> CH3COONa + H2O}\)

Given the two equations above. What volume of the same sodium hydroxide solution would be required to completely react with 25.0 mL of a 0.100 mol L\(^{-1}\) acetic acid solution \(\ce{(CH3COOH)}\)?

  1. Less than  23.4 mL
  2. 23.4 mL
  3. More than  23.4 mL
  4. Unable to calculate unless the concentration of the sodium hydroxide solution is also known
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\(B\)

Show Worked Solution

→ The strength of the acid does not affect the volume of \(\ce{NaOH}\) required for nuetralisation.

→ It is dependent on stoichiometric ratios and both \(\ce{HCl}\) and \(\ce{CH3COOH}\) are monoprotic and so react in 1:1 ratio with sodium hydroxide.

\(\Rightarrow B\)