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Chemistry

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4 parts
a

1 mol\textbf{1 mol}

b

29.25 g\textbf{29.25 g}

c

4.48 dm3\textbf{4.48 dm}^3

d

0.1 mol\textbf{0.1 mol}

Step-by-step solution

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Part a· 2 steps

Calculate the number of moles of 18 g18 \text{ g} of water.

  1. Determine the molar mass of water (H2O\text{H}_2\text{O}).

    The atomic mass of hydrogen (H\text{H}) is 1 g/mol1 \text{ g/mol}.
    The atomic mass of oxygen (O\text{O}) is 16 g/mol16 \text{ g/mol}.

    Molar mass of H2O=(2×1 g/mol)+(1×16 g/mol)=2 g/mol+16 g/mol=18 g/mol\text{Molar mass of H}_2\text{O} = (2 \times 1 \text{ g/mol}) + (1 \times 16 \text{ g/mol}) = 2 \text{ g/mol} + 16 \text{ g/mol} = 18 \text{ g/mol}

  2. Calculate the number of moles.

    Moles=MassMolar mass\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} Moles=18 g18 g/mol=1 mol\text{Moles} = \frac{18 \text{ g}}{18 \text{ g/mol}} = 1 \text{ mol}

    The number of moles of 18 g18 \text{ g} of water is 1 mol\boxed{\textbf{1 mol}}.

Part b

29.25 g\textbf{29.25 g}

Calculate the mass of 0.5 moles0.5 \text{ moles} of sodium chloride.

  1. Determine the molar mass of sodium chloride (NaCl\text{NaCl}).

    The atomic mass of sodium (Na\text{Na}) is 23 g/mol23 \text{ g/mol}.
    The atomic mass of chlorine (Cl\text{Cl}) is 35.5 g/mol35.5 \text{ g/mol}.

    Molar mass of NaCl=23 g/mol+35.5 g/mol=58.5 g/mol\text{Molar mass of NaCl} = 23 \text{ g/mol} + 35.5 \text{ g/mol} = 58.5 \text{ g/mol}

  2. Calculate the mass.

    Mass=Moles×Molar mass\text{Mass} = \text{Moles} \times \text{Molar mass} Mass=0.5 mol×58.5 g/mol=29.25 g\text{Mass} = 0.5 \text{ mol} \times 58.5 \text{ g/mol} = 29.25 \text{ g}

    The mass of 0.5 moles0.5 \text{ moles} of sodium chloride is 29.25 g\boxed{\textbf{29.25 g}}.

Part c

4.48 dm3\textbf{4.48 dm}^3

Calculate the volume of 0.2 moles0.2 \text{ moles} of oxygen gas at STP.

  1. State the molar volume of any gas at Standard Temperature and Pressure (STP).

    At STP, 1 mole1 \text{ mole} of any gas occupies 22.4 dm322.4 \text{ dm}^3.

  2. Calculate the volume.

    Volume=Moles×Molar volume at STP\text{Volume} = \text{Moles} \times \text{Molar volume at STP} Volume=0.2 mol×22.4 dm3/mol=4.48 dm3\text{Volume} = 0.2 \text{ mol} \times 22.4 \text{ dm}^3\text{/mol} = 4.48 \text{ dm}^3

    The volume of 0.2 moles0.2 \text{ moles} of oxygen gas at STP is 4.48 dm3\boxed{\textbf{4.48 dm}^3}.

Part d

0.1 mol\textbf{0.1 mol}

Calculate the number of moles of 2.24 dm32.24 \text{ dm}^3 of carbon dioxide gas at STP.

  1. State the molar volume of any gas at Standard Temperature and Pressure (STP).

    At STP, 1 mole1 \text{ mole} of any gas occupies 22.4 dm322.4 \text{ dm}^3.

  2. Calculate the number of moles.

    Moles=VolumeMolar volume at STP\text{Moles} = \frac{\text{Volume}}{\text{Molar volume at STP}} Moles=2.24 dm322.4 dm3/mol=0.1 mol\text{Moles} = \frac{2.24 \text{ dm}^3}{22.4 \text{ dm}^3\text{/mol}} = 0.1 \text{ mol}

    The number of moles of 2.24 dm32.24 \text{ dm}^3 of carbon dioxide gas at STP is 0.1 mol\boxed{\textbf{0.1 mol}}.

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