CAIE A-Level Chemistry 4. States of matter Question Bank

Practise gas calculations and state-of-matter explanations using pV = nRT, ideal behaviour, volatility and bonding models.

[Maximum number: 3]

Nitrogen and phosphorus are elements in Group 15 of the Periodic Table.

(a)

Complete Table 3.1 to show the properties of nitrogen and phosphorus in their standard states.

Table 3.1

[ 2 ]

(b)

A form of solid nitrogen has a lattice structure similar to solid iodine.

Identify the type of lattice structure of solid nitrogen.

[ 1 ]

[Maximum number: 7]

The strength of interaction between particles determines whether the substance is a solid, liquid or gas at room temperature.

(a)

Lithium sulfide, $\mathrm{Li}_{2} \mathrm{~S}$ , is a crystalline solid with a melting point of $938^{\circ} \mathrm{C}$ . It conducts electricity when it is molten.

[ 3 ]

(i)

Give the formulae of the particles present in solid lithium sulfide.

[ 1 ]

$\mathrm{Li}^{+}$ AND S ${ }^{2-}$

(ii)

Explain, in terms of the structure of the crystalline solid, why lithium sulfide has a high melting point.

[ 2 ]

(b)

Nitrogen, $\mathrm{N}_{2}$ , is also a gas at room temperature and pressure. Neither CO nor $\mathrm{N}_{2}$ is an ideal gas.

[ 4 ]

(i)

State two assumptions that are made about the behaviour of particles in an ideal gas.

1
2

[ 2 ]

(ii)

Explain why $\mathrm{N}_{2}$ does not behave as an ideal gas at very high pressures.

[ 2 ]

[Maximum number: 3]

The Pauling electronegativity values of elements can be used to predict the chemical properties of compounds.

Use the information in Table 1.1 to answer the following questions.

Table 1.1

At $150^{\circ} \mathrm{C}$ and 103 kPa , all of the compounds listed in Table 1.2 are gases.

Under these conditions, 0.284 g of one of the compounds occupies a volume of $127 \mathrm{~cm}^{3}$ .
Use this information to calculate the $M_{\mathrm{r}}$ of the compound. Hence, identify the compound from those given in Table 1.2.

Show your working.
$M_{\mathrm{r}}=$ identity of compound =

conversion of units

$103000 \mathrm{~Pa} \quad 127 \times 10^{-6} \mathrm{~m}^{3}$

Use of $p V=\left(m / M_{\mathrm{r}}\right) R T$ $M_{\mathrm{r}}=\frac{0.284 \times 8.31 \times 423}{103000 \times 127 \times 10^{-6}}$ $M_{\mathrm{r}}=76.3$ AND compound $=\mathrm{CS}_{2}$

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Covered syllabus points

4. States of matter question 1

Question (a)

Question (b)

4. States of matter question 2

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)

Question (ii)

4. States of matter question 3