Some Basic Concepts of Chemistry

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(i) → (e) (ii) → (d) (iii) → (b) (iv) → (g) (v) → (c), (h) (vi) → (f) (vii) → (a) (viii) → (i)

Explanation:

(i) It is defined as the number of moles of solute dissolved in 1 litre of solution. So, the unit of molarity is mol L-1
(ii) M

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(i)- (b); (ii) - (c); (iii) - (a); (iv) - (e); (v) - (d)

(i) 88 g of CO₂

The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  - (1)

So, the number of moles of CO₂ is calculated by using equation (1) as follows

Moles of

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Option (A), (D)

Law of conservation of mass: According to this law, the mass is neither created nor destroyed during a chemical reaction in an isolated system. It can only be transformed from one form to another. Burning of wood is an

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Option (C), (D)

Molality: It is defined as the number of moles of solute divided by the mass of solvent in kg . It is denoted by m . The formula is expressed as

m = $\frac{\mathrm{n}}{\mathrm{W}\mathrm{}\mathrm{}}$

The SI unit of molality is 1 mol/kg

Molarity: It is defined as the num

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Option(C), (D)

The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  ---------------(1)

The number of molecules can be calculated as, number of moles

= $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{A}\mathrm{v}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{}\mathrm{s}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}}$  ----------(2)

(C) The number of moles of N₂ is calculated by

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The answers are options (i) and (ii)

(A)The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

So, the number of moles of NaOH is calculated, by using equation (1) as follows

 20 g NaOH in 200 mL solution.

Moles of NaOH=

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Option(C), (D)

(A) The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}\mathrm{}$ -----------(1)

The number of moles of O₂ is calculated by using equation (1) as follows,

Moles of O₂ = $\frac{16\mathrm{}\mathrm{g}}{32\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  = 0.5 mol

The number of atoms can be calcul

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Option (B), (C)

0.1 mole H₂SO₄ reacts with 1 mole of NaOH.

0.1 mole of NaOH will react with $\frac{0.1}{2\mathrm{}\mathrm{}}$ = mole of H₂SO₄

Here, NaOH is the limiting reagent.

2 mole of NaOH produces 1 mole of Na₂SO₄

0.1 mole of NaOH will give $\frac{0.1}{2\mathrm{}\mathrm{}}$  mole of Na₂SO₄

No

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Option (A), (D)

One mole of oxygen gas at STP is equal to  6.022 x 10²³  molecules of oxygen The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}}$    ----------(1)

The number of moles of O₂ is calcula

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Option (B)

Law of multiple proportion: According to the law of multiple proportion, whe form two or more than two chemical compounds, the ratio between different elements combining with a fixed mass of the other is always in the ratio