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Define the term allotropy and name the two allotropes of sulphur.

Chemistry

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Answer

b

18.4\mol\dm−318.4 \mol\dm^{-3}

Step-by-step solution

4 steps

Part a

Allotropy is the existence of two or more different physical forms (allotropes) of the same element in the same physical state. These forms differ in physical properties but have similar chemical properties.

Part b

18.4\mol\dm−318.4 \mol\dm^{-3}

The two allotropes of sulphur are:
rhombic sulphur and monoclinic sulphur.

9
Concentrated sulphuric(VI) acid (\ceH2SO4\ce{H2SO4}) is 98%98\% by mass and has a density of 1.84 g cm−3^{-3}. Calculate its molarity.

Molar mass of \ceH2SO4=2(1)+32+4(16)=98\ce{H2SO4} = 2(1) + 32 + 4(16) = 98 g mol−1^{-1}.

  1. Mass of 1 dm3^3 (1000 cm3^3) of solution.

    Mass=density×volume=1.84 g cm−3×1000 cm3=1840 g.\text{Mass} = \text{density} \times \text{volume} = 1.84 \, \text{g cm}^{-3} \times 1000 \, \text{cm}^3 = 1840 \, \text{g}.

  2. Mass of \ceH2SO4\ce{H2SO4} in 1 dm3^3 of solution.

    Mass of \ceH2SO4=98%×1840 g=0.98×1840=1803.2 g.\text{Mass of } \ce{H2SO4} = 98\% \times 1840 \, \text{g} = 0.98 \times 1840 = 1803.2 \, \text{g}.

  3. Moles of \ceH2SO4\ce{H2SO4}.

    Moles=1803.2 g98 g mol−1=18.4 mol.\text{Moles} = \frac{1803.2 \, \text{g}}{98 \, \text{g mol}^{-1}} = 18.4 \, \text{mol}.

  4. Molarity.

    Molarity=18.4 mol1 dm3=∗∗18.4 mol dm−3∗∗.\text{Molarity} = \frac{18.4 \, \text{mol}}{1 \, \text{dm}^3} = **18.4 \, \text{mol dm}^{-3}**.

    18.4 \mol\dm^{-3}

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