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Question

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Explanation / Answer

18.

The given chemical equation which is to be balanced is,

S + HNO3 = H2SO4 + NO2 + H2O

The chemical equation after balancing is,

S + 6HNO3 ---------------> H2SO4 + 6NO2 + 2H2O

In this balanced equation, the coefficient of water is 2.

So, option E is correct.

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19.

Law of definite proportion states that a chemical compound always contains exactly the same proportion of elements by mass. This is also called Law of constant composition.

Given, a chemical compound X contains, 5.0 g Oxygen, 10.0 g Carbon and 20.0 g Nitrogen.

So, it means that 10.0 g of Carbon is present in 35.0 g of total chemical X.

So, the percentage proportion of carbon is (10.0 / 35.0) x 100 = 28.5 %

And, we were asked to find the amount in 71 g of sample of Chemical X.

It means that, 28.5% of 71 g sample of chemical X is Carbon.

Amount of Carbon = 71 x (28.5 / 100) = 20.2 =20 g

So, option B is correct.

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20.

From the solubility rules, we know that, maximum of all Nitrates are soluble.

That means, it has fewest insoluble salts.

So, option B is correct.

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21.

Iodine in option A (I2) is zero, as it is in Molecular form.

In NH4I, N has -3, H has +1, and I has -1 oxidation state.

In LiIO3, Li has +1, I has +5 , O has -2 oxidation states.

In IO2, I has +4, and O has -2 oxidation states.

In I2O, I has+1 and O has -2 oxidation states.

So, only NH4I has I with -1 oxidation state, which is less among the given.

So, option B is correct.

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22.

Given that the isotope has 23 protons.

It means that, the atomic number of the element is 23.

The element with atomic number 23 is Vanadium [V].

So, option C or E is correct.

And it was given that there are 18 electrons.

So, for option C, 50 V 5-, electrons = Protons - (charge)

Electrons = 23-(-5) = 28 (Incorrect)

For option E, 50 V 5+, electrons = Protons - (charge)

Electrons = 23-(+5) = 18 (Correct)

Let us check for neutrons finally.

Neutrons = Mass number - Protons

= 50-23 = 27

Option E satisfies all the criteria.

So, option E is correct.

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23.

Given equation is AgNO3 + LiI --------------> AgI + LiNO3

AgI is a precipitate.

So, the reaction is AgNO3(aq) + LiI(aq) --------------> AgI(s) + LiNO3(aq)

So, the spectator ions are Li+ and NO3-

So,option B is correct.

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