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CHEMISTRY, M2 EQ-Bank 10 MC

Chlorine is made up of two isotopes with mass numbers 35 and 37.

Determine the percentages of the two isotopes of chlorine given the relative atomic mass is 35.45.

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Chlorine-35}\rule[-1ex]{0pt}{0pt}& \text{Chlorine-37} \\
\hline
\rule{0pt}{2.5ex}\text{47.5%}\rule[-1ex]{0pt}{0pt}&\text{52.5%}\\
\hline
\rule{0pt}{2.5ex}\text{57.5%}\rule[-1ex]{0pt}{0pt}& \text{42.5%}\\
\hline
\rule{0pt}{2.5ex}\text{67.5%}\rule[-1ex]{0pt}{0pt}& \text{32.5%} \\
\hline
\rule{0pt}{2.5ex}\text{77.5%}\rule[-1ex]{0pt}{0pt}& \text{22.5%} \\
\hline
\end{array}
\end{align*}

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\(D\)

Show Worked Solution
  • Let \(x\) be the percentage of chlorine-35 present, therefore the percentage of chlorine-37 present is \(100-x\).
\(\dfrac{35x +37(100-x)}{100}\) \(=35.45\)  
\(3700-2x\) \(=3545\)  
\(2x\) \(=155\)  
\(x\) \(=77.5\)  

 
\(\Rightarrow D\)

CHEMISTRY, M2 EQ-Bank 5 MC

In which molecule is the percentage by mass of oxygen closest to 70%?

  1. \(\ce{H2O}\)
  2. \(\ce{SO2}\)
  3. \(\ce{CO2}\)
  4. \(\ce{CH2O}\)
Show Answers Only

\(C\)

Show Worked Solution
  • For \(\ce{H2O}\): \(MM = 2(1.008) + 16.00 = 18.016\ \text{g mol}^{-1}\).
  •    % Mass of oxygen \(=\dfrac{16.00}{18.016} \times 100 = 88.8\)%. 
  • For \(\ce{SO2}\): \(MM = 32.07 + 2(16.00) = 64.07\ \text{g mol}^{-1}\).
  •    % Mass of oxygen \(=\dfrac{2(16.00)}{64.07} \times 100 = 49.9\)%.
  • For \(\ce{CO2}\): \(MM = 12.01 + 2(16.00) = 44.01\ \text{g mol}^{-1}\).
  •    % Mass of oxygen \(=\dfrac{2(16.00)}{44.01} \times 100 = 72.7\)%.
  • For \(\ce{CH2O}\): \(MM = 12.01 + 2(1.008) + 16.00 = 30.026\ \text{g mol}^{-1}\).
  •    % Mass of oxygen \(=\dfrac{16.00}{30.026} \times 100 = 53.3\)%.

\(\therefore \ce{CO2}\) has the closest oxygen percentage to 70%.

\(\Rightarrow C\)

CHEMISTRY, M2 EQ-Bank 2 MC

What would the percentage of calcium in a sample of calcium carbonate \(\ce{(CaCO3)}\) be closest to?

  1. 30 %
  2. 40 %
  3. 50 %
  4. 60 %
Show Answers Only

\(B\)

Show Worked Solution

\(\ce{MM(CaCO3)} = 40.08 + 12.01 + 3(16.00) = 100.09\ \text{g mol}^{-1}\).

  • The mass of calcium is \(40.08\ \text{g mol}^{-1}\).
  • Percentage composition of calcium in \(\ce{(CaCO3)} = \dfrac{40.08}{100.09} \times 100 \approx 40\)%.

\(\Rightarrow B\)

CHEMISTRY, M2 2005 HSC 25a

A student collected a 500 mL sample of water from a local creek for analysis. It was filtered and the filtrate evaporated to dryness. The following data were collected.

\begin{array} {|l|r|}
\hline
\rule{0pt}{2.5ex}\text{Mass of filter paper}\rule[-1ex]{0pt}{0pt} & \text{0.16 g}\\
\hline
\rule{0pt}{2.5ex}\text{Mass of filter paper and solid}\rule[-1ex]{0pt}{0pt} & \text{0.19 g}\\
\hline
\rule{0pt}{2.5ex}\text{Mass of evaporating basin}\rule[-1ex]{0pt}{0pt} & \text{45.33 g}\\
\hline
\rule{0pt}{2.5ex}\text{Mass of basin and solid remaining}\rule[-1ex]{0pt}{0pt} & \text{45.59 g}\\
\hline
\end{array}

Calculate the percentage of total dissolved solids in the creek sample.  (2 marks)

Show Answers Only

\(89.66\% \)

Show Worked Solution
  • The dissolved solids in the creek sample are contained in the evaporating basin and the undissolved solids are contained in the filter paper.
  • Mass of undissolved solids in filter paper = \(0.19-0.16 = 0.03\ \text{g}\)
  • Mass of dissolved solids in evaporating basin = \(45.59-45.33 = 0.26\ \text{g}\)
  • Total Mass of solids = \(0.29\ \text{g}\)
  • Percentage of dissolved solids = \(\dfrac{0.26}{0.29} \times 100 = 89.66\% \) of total solids in the creek sample

CHEMISTRY, M2 2009 HSC 22

The nitrogen content of bread was determined using the following procedure:

    • A sample of bread weighing 2.80 g was analysed.
    • The nitrogen in the sample was converted into ammonia.
    • The ammonia was collected in 50.0 mL of 0.125 mol L\(^{-1}\) hydrochloric acid. All of the ammonia was neutralised, leaving an excess of hydrochloric acid.
    • The excess hydrochloric acid was titrated with 23.30 mL of 0.116 mol L\(^{-1}\) sodium hydroxide solution.
  1. Write balanced equations for the TWO reactions involving hydrochloric acid.   (2 marks)

  2. Calculate the moles of excess hydrochloric acid.   (1 mark)

  3. Calculate the moles of ammonia.   (2 marks)

  4. Calculate the percentage by mass of nitrogen in the bread.   (2 marks)

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a.    \(\ce{NH3(g) + HCl(aq) \rightarrow NH4+(aq) + Cl-(aq)}\)

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

b.   \(2.70 \times 10^{-3}\)

c.   \(3.55 \times 10^{-3}\)

d.   \(1.78\%\)

Show Worked Solution

a.    \(\ce{NH3(g) + HCl(aq) \rightarrow NH4+(aq) + Cl-(aq)}\)

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

b.    \(\ce{n(NaOH excess) = c \times V = 0.116 \times 0.02330 = 2.70 \times 10^{-3} moles}\)
 

c.    \(\ce{n(HCl original) = c \times V = 0.125 \times 0.0500 = 6.25 \times 10^{-3} moles}\)

\(\ce{n(HCl used) = 6.25 \times 10^{-3}-2.70 \times 10^{-3} = 3.55 \times 10^{-3} moles}\)

\(\ce{n(NH3) = n(HCl used) = 3.55 \times 10^{-3} moles}\)
 

d.    \(\ce{n(N) = n(NH3) = 3.55 \times 10^{-3} moles}\)

\(\ce{Mass N = 3.55 \times 10^{-3} \times 14.01 = 0.0497 g}\)

\[\ce{\% N (by mass) = \frac{0.0497}{2.80} \times 100\% = 1.78\%}\]

CHEMISTRY, M2 2010 HSC 31ai

A student collected a 250 mL sample of water from a local dam for analysis. The data collected are shown in the table.

\begin{array} {|l|r|}
\hline
\rule{0pt}{2.5ex}\text{Mass of filter paper}\rule[-1ex]{0pt}{0pt} & \text{0.23 g} \\
\hline
\rule{0pt}{2.5ex}\text{Mass of filter paper and solid}\rule[-1ex]{0pt}{0pt} & \text{0.47 g} \\
\hline
\rule{0pt}{2.5ex}\text{Mass of evaporating basin}\rule[-1ex]{0pt}{0pt} & \text{43.53 g} \\
\hline
\rule{0pt}{2.5ex}\text{Mass of basin and solid remaining}\ \ \rule[-1ex]{0pt}{0pt} & \text{44.67 g} \\
\hline
\end{array}

The water was filtered and the filtrate evaporated to dryness.

Calculate the percentage of the total dissolved solids in the dam sample.  (2 marks)

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0.46%

Show Worked Solution

\(\text{Total Dissolved Solids}\ = 44.67-43.53 = 1.14\ \text{g}\)

\[\text{% TDS in sample}\ = \frac{1.14}{250} \times 100\% = 0.46\%\ \text{(to 2 d.p.)}\]