Molar mass $$NH_3$$ = 14+3 = 17g/mol 

Given mass = 4.25g

No of moles= $$\dfrac{given\ mass}{molar\ mass}$$

No of moles = 4.25/17 = 0.25 mol of  $$NH_3$$

1 mol of  $$NH_3$$ contains  $$6.022\times 10^{23}$$  molecules 

0.25 mol $$NH_3$$ contains  $$0.25 \times$$$$6.022\times 10^{23}$$  = $$1.5055\times$$ $$10^{23}$$ molecules 

Since the percentage of $$P$$ is $$0.086$$

So $$1$$ mole of oxygen $$=16\ grams$$

No of Molecules = $$\dfrac{Weight}{Gram\space Molecular\space Weight} \times {N}_{A}$$

Molar mass of $$Na_2CO_3$$ $$= (2 \times atomic\ mass\ of\ nitrogen) + (1 \times atomic\ mass\ of\ carbon) + (3 \times atomic\ mass\ of\ oxygen)$$ $$= 2 (23) + 1 (12) + 3 (16)$$ $$= 106\ g/mol$$

Atomic mass of $$H=1$$

Molecules containing more than 3 atoms are called polyatomic molecules.

Moles of $$CaCO_3=\cfrac {100}{10^3 \times 100}=1 \ milli moles$$

Let the atomic mass of $$A$$ be $$M_A$$ and $$B$$ be $$M_B$$

Hint: Mass of an atom is directly proportional to number of moles.

$$ \begin{array}{l} 22 \times 10^{-3} \text {mg per year }=22 \times 10^{-6} \mathrm{~g} \text { peryear } \\ =\frac{22 \times 10^{-6} \mathrm{~g}}{365 \times 24 \times 3600} \\ =6.976 \times 10^{-13} \mathrm{~g} \text { persecond } \\ \text { Thus, mass of } 1.06 \times 10^{11} \alpha \text { particles. } \\ \qquad \begin{aligned} =6.976 \times 10^{-13} \mathrm{~g} \\ \text {  } \end{aligned} \end{array} $$

The minimum number of oxygen atoms the compound can contain is $$1.$$

1 mole of $$C_{2}H_{2}O_{4} \longrightarrow$$2 moles of hydrogen=2 g of hydrogen

The percentage of Iron is $$4.6 \%$$

$$Na_2SO_4.10H_2O$$

Atomic weight of $$O=16\ g/mole$$

$$pV=nRT$$

          The law of definite proportion or law of constant composition states that a given chemical compound always contains its component elements in a fixed ratio (by mass) and does not depend on its source and method of preparation.

Given,

Number of molecules of $$X = 3\times 10^{21}$$

Number of molecules of $$Y = 4.5\times 10^{25}$$

Molecular weight of $$X = 50$$

Total mass of mixture $$= 1\ g$$ 

Formula used :

$$Mass\ of\ mixture=\left[\dfrac{Number\ of\ molecules\ of\ X}{Avogrado's\ Number} \times Molar\ Mass\ of\ X\right]$$

                                 $$+\left[\dfrac{Number\ of\ molecules\ of\ Y}{Avogrado's\ Number}\times  Molar\ Mass\ of\ Y\right]$$

Now, putting all the given values in this formula, we get the molecular mass of $$Y$$.

$$1=\left[\dfrac{3\times 10^{21}}{6.023\times 10^{23}}\times 50\right]+\left[\dfrac{4.5\times 10^{25}}{6.023\times 10^{23}}\times Y\right]$$

$$\Rightarrow 1=\dfrac{150\times 10^{21}+4.5Y\times 10^{25}}{6.023\times 10^{23}}$$

$$\Rightarrow 6.023\times 10^{23}=10^{21}(150+4.5Y\times 10^4)$$

$$\Rightarrow \dfrac{6.023\times 10^{23}}{10^{21}}=150+4500Y$$

$$\Rightarrow 602.3=150+4500Y$$

$$\Rightarrow 4500Y=602.3-150$$

$$\Rightarrow Y = \dfrac{450.3}{4500}$$

$$Y= 150$$

Therefore, on solving, we get $$Y=150$$

Therefore, the molecular weight of $$Y$$ is $$150g$$.

Hence option (B) is correct.

$$1\ mol$$ of $$Na_{2}B_{4}O_{4} \cdot 10H_{2}O$$ contains $$6.022\times 10^{23} molecules$$

Here, Avogadro number of helium atoms means, $$6.023 \times 10^{23}$$ atoms of He.

The word element was first given by Rebert Boyle. It is defined as the simplest chemical substance that cannot be broken down during a chemical reaction.

Hint: The minimum number of Sulphur atoms which a molecule can possess is one.

Step 1: Finding the molecular mass of the compound.

$$Mass\ of\ 1\ sulphur\ atom\ =\ 32 \ grams.$$

$$ As\ the\ compound\ has\ 8\% \ sulphur\ by\ mass,$$

$$\therefore 8 \ grams\ of\ sulphur\ is\ present\ in\ 100 \ grams\ of\ compound.$$

$$\therefore 32 \ grams\ of\ sulphur\ is\ present\ in\ \dfrac{32}{8} \times 100=400\ grams \ of\ compound.$$

$$Hence,\ the\ least\ molecular\ mass\ of\ the\ compound\ is\ 400 \ grams.$$

Final Step: Correct answer - (C) The least molecular mass of the compound is $$400 \ grams.$$

Accepted as unit of atomic and molecular mass is atomic mass unit (u).