smc-4262-40-Volume

CHEMISTRY, M2 2009 HSC 15 MC

The graph shows the maximum dissolved oxygen concentration in water as a function of temperature at normal atmospheric pressure.

What is the volume of \(\ce{O2}\) that can dissolve in 10.0 L of water at 25°C and normal atmospheric pressure?

62.0 mL
63.5 mL
80.0 mL
124 mL

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\(A\)

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Concentration of \(\ce{O2}\) at 25°C is 8 \(\text{mgL}^{-1}\), thus in 10 \(\text{L}\), 80 \(\text{mg}\).
\(n(\ce{O2})=\dfrac{80 \times 10^{-3}}{32}= 0.0025\ \text{mol}\)
\(v(\ce{O2})= 0.0025 \times 24.79= 0.0620\ \text{L} = 62.0\ \text{mL}\)

\(\Rightarrow A\)

CHEMISTRY, M2 2010 HSC 26

A gas is produced when 10.0 g of zinc is placed in 0.50 L of 0.20 mol L\(^ {-1}\) nitric acid.

Calculate the volume of gas produced at 25°C and 100 kPa. Include a balanced chemical equation in your answer. (4 marks)

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\(1.24\ \text{L}\)

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\(\ce{2HNO3 + Zn \rightarrow Zn(NO3)2 + H2}\)

\(\ce{n(HNO3) = c \times V = 0.20 \times 0.5 = 0.10 moles}\)

\(\ce{n(Zn) = \dfrac{\text{m}}{\ce{MM}} = \dfrac{10}{65.38} = 0.153 moles}\)

\(\ce{HNO3}\) and \(\ce{Zn}\) react in a \(2:1\) ratio
\(\ce{HNO3}\) is the limiting reagent and 0.05 moles of gas \(\ce{H2}\) will be produced
\(\ce{V(H2) = 0.05 \times 24.79 = 1.24\ \text{L (2 d.p.)}}\)

CHEMISTRY, M2 2011 HSC 19 MC

All of the carbon dioxide in a soft drink with an initial mass of 381.04 g was carefully extracted and collected as a gas. The final mass of the drink was 380.41 g.

What volume would the carbon dioxide occupy at 100 kPa and 25°C?

0.33 L
0.35 L
0.56 L
0.63 L

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\(B\)

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\(\ce{m(CO2)}= 0.63\ \text{g}\)

\(\ce{n(CO2)}= \dfrac{0.63}{44.01}=0.0143\ \text{mol}\)

\(\ce{V(CO2)}= 0.0143 \times 24.79=0.35\ \text{L}\)

\(\Rightarrow B\)

CHEMISTRY, M2 2013 HSC 14 MC

Sodium reacts with water to give hydrogen gas and sodium hydroxide solution.

What volume of gas would be produced from the reaction of 22.99 g of sodium at 25°C and 100 kPa ?

11.36 L
12.40 L
22.71 L
24.79 L

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\(B\)

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\(\ce{2Na(s) + 2H2O(l) \rightarrow H2(g) + 2NaOH(aq)}\)

\(n(\ce{Na(s)}) = \dfrac{22.99}{22.99}= 1\ \text{mol}\)

\(n(\ce{H2(g)}) = \dfrac{1}{2} \times 1= 0.5\ \text{mol}\)

The volume of \(\ce{H2(g)}= 0.5 \times 24.79 = 12.40\ \text{L}\)

\(\Rightarrow B\)

An unattended car is stationary with its engine running in a closed workshop. The workshop is 5.0 m × 5.0 m × 4.0 m and its volume is \(1.0\ ×\ 10^{5}\) L. The engine of the car is producing carbon monoxide in an incomplete combustion according to the following chemical equation:

\(\ce{C8H18(l)+\dfrac{17}{2}O2(g) \rightarrow 8CO(g) + 9H2O(l)}\)

Exposure to carbon monoxide at levels greater than 0.100 g L\(^{-1}\) of air can be dangerous to human health.

6.0 kg of octane was combusted by the car in this workshop.

Using the equation provided, determine if the level of carbon monoxide produced in the workshop would be dangerous to human health. Support your answer with relevant calculations. (4 marks)

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\(\ce{\text{Volume of garage} = 5 \times 5 \times 4 = 100 \text{m}^2}\)

\(\ce{100 \text{m}^2} = 100\ 000\ \text{L}\)

\[\ce{\text{n(octane)} = \frac{n}{M} = \frac{6000}{114.224} = 52.53 \text{moles}} \]

\(\ce{Molar ratio of Octane : CO = 1:8}\)

\(\ce{n(CO) = 8 \times 52.53 = 420.23}\)

\(\ce{m(CO) = n \times M = 420.23 \times 28.01 = 11\ 771\ g}\)

\[\ce{[CO] = \frac{11\ 771}{100\ 000} = 0.118 g L^{-1}}\]

\(\ce{\text{Since} [CO] > 0.100 g L^{-1}, \text{it is dangerous to human health.}}\)

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\(\ce{\text{Volume of garage} = 5 \times 5 \times 4 = 100 \text{m}^2}\)

\(\ce{100 \text{m}^2} = 100\ 000\ \text{L}\)

\[\ce{\text{n(octane)} = \frac{n}{M} = \frac{6000}{114.224} = 52.53 \text{moles}} \]

\(\ce{Molar ratio of Octane : CO = 1:8}\)

\(\ce{n(CO) = 8 \times 52.53 = 420.23}\)

\(\ce{m(CO) = n \times M = 420.23 \times 28.01 = 11\ 771\ g}\)

\[\ce{[CO] = \frac{11\ 771}{100\ 000} = 0.118 g L^{-1}}\]

\(\ce{\text{Since} [CO] > 0.100 g L^{-1}, \text{it is dangerous to human health.}}\)