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151.

Name the particle which has 18 electrons , 18 neutrons and 17 protons in it.

Answer» Correct Answer - Chlorine ion, `Cl^(-)`
Discussion
152.

What is the name of a particle which contains 10 electrons, 11 protons and 12 neutrons?

Answer» Correct Answer - Sodium ion, `Na^(+)`
Discussion
153.

In an experiment, `1.288g` of copper oxide was obtained from `1.03 g` of copper. In another experiment `3.672` g of copper oxide gave, on reduction, `2.938` g of copper. Show that these figures verify the law of constant proportions.

Answer» In order to solve this problem we have to calculate the ratio (or proportion) of copper and oxygen in the two samples of copper oxide compound. Now,
(a) In the first experiment:
Mass of copper `= 1.03 g "….."(1)`
And, Mass of copper oxide `= 1.288 g`
So, Mass of oxygen = Mass of copper oxide - Mass of copper
`= 1.288-1.03"......"(2)`
Now, in the first sample of copper oxide compound:
Mass of copper: Mass of oxygen `= 1.03 : 0.258`
`= (1.03)/(0.258) : 1`
`= 3.99: 1`
`= 4 : 1"......."(3)`
(b) In the second experiment:
Mass of copper = 2.938 g`"......"(4)`
And, Mass of copper oxide `= 3.672 - 2.938`
`= 0.734 g "........"(5)`
Now, in the second sample of copper oxide compound:
Mass of copper : Mass of oxygen `= 2.938 : 0.734`
`= (2.938)/(0.734) : 1`
`= 4 : 1"........"(6)`
From the above calculations we can see that the ratio (or proportion) of copper and oxygen elements in the two samples of copper oxide compound is the same `4 : 1`. So, the given figures verify the law of constant proportions.
Discussion
154.

Calculate the number of molecules of sulphur `(S_(8))` present in 16 g of solid sulphur.

Answer» The molecular formula of sulphur is given to the be ` S_(8)` (it contains 8 atoms of sulphur). So, the molecular mass of sulphur molecular is `32 xx 8 = 256 u`. This means that 1 mole of sulphur molecules is equal to `256` grams.
Now, `256` g of sulphur = 1 mole of sulphur molecules
So, 16 g of sulphur ` = 1/256 xx 16` mole of sulphur molecules
`= 0.625` mole of sulphur molecules
Also, 1 mole of sulphur molecules `= 6.023 xx 10^(23)` molecules
So, `0.0625` mole of sulphur molecules `= 6.023 xx 10^(23) xx 0.0625` molecules
`= 3.76 xx 10^(22)` molecules
Discussion
155.

In an experiment, 4.90 g of copper oxide was obtained from 3.92 g of copper. In another experiment, 4.55 g of copper oxide gave, on reduction, 3.64 g of copper. Show with the help of calculation that these figures verify the law of constant proportions.

Answer» Correct Answer - The ratio (or proportion) of copper and oxygen elements in the two samples of copper oxide compound is the same `4 : 1`. So, the given figures verify the law of constant proportions
Discussion
156.

Work out the formulae for the following compounds : (a) Sodium oxide , (b) Calcium carbonate

Answer» Correct Answer - (a) `Na_(2)O` , (b) `CaCO_(3)`
Discussion
157.

Give the formulae of the compounds formed from the following sets of elements. a) Calcium and fluroine b) Hydrogen and sulphur c) Nitrogen and hydrogen d) Carbon and chlorine e) Sodium and oxygen f) Carbon and oxygen

Answer» `(a)CaF_(2) (b)H_(2)S (c)NH_(3) (d)C Cl_(2) (e)Na_(2)O (f)CO and CO_(2)`
Discussion
158.

Give the formulae of the compounds that will be formed from the following sets of elements. (a) Calcium and fluorine (b) Magnesium and oxygen (c )Sodium and sulphur (d) Carbon and chlorine (e )Carbon and sulphur (f) Nitrogen and hydrogen.

Answer» `(a)CaF_(2)` (b)`MgO` (c )`Na_(2)S`(d` C Cl_(4)` (e)`CS_(2)` (f) `NH_(3)`
Discussion
159.

Write the chemical formulae of the following: (a)Magnesium chloride (b)Calcium oxide (c )Sodium sulphide (d)Aluminium phosphate (e)potassium chloride (f)Calcium carbonate.

Answer» `(a) MgSO_(4), (b) CaO (c) Na_(2)S (d) Al_(2)(PO_(4))_(3), (e) KCI (f) CaCO_(3)`.
Discussion
160.

Write down the formulae of (i) sodium oxide (ii) aluminium chloride (iii) sodium sulphide (iv) magnesium hydroxide

Answer» `Na_(2)O`
`AlCl_(3)`
`Na_(2)S`
`Mg(OH)_(2)`
Discussion
161.

Calcium chlroide when dissolved in water dissociates into its ions according to the following equations. `CaCl_(2)(aq) to Ca^(2+)(aq) + 2Cl^(-1) (aq)` Calculate the number of ions obtained from `CaCl_(2)` when 222g of it is dissolved in water.

Answer» Molar mass of `CaCl_(2)=40+(2xx35.5)=111"g mol"^(-1)`
`111"g of "CaCl_(2)"represent"=1"mol"`
`222"g of "CaCl_(2)"represent"=((222g))/((111g))xx1"mol"=2"mol"`
No. of molecules in 2 moles of `CaCl_(2)=2xxN_(A)=2xx6.022xx10^(23)`
1 molecules of `CaCl_(2)` in solution form ions =3
`2xx6.022xx10^(23)` molecules of `CaCl_(2)` in solution from ions `=3xx2xx6.022xx10^(23)`
`=36.132xx10^(23)`
`=3.6132xx10^(24)`
Discussion