Question

which of the following has maximum number of atoms a 18 gram of H2O option B 18 gram of O2 option C 18 gram of co2 option d 18 gram of ch4​

Answer 1

Heyy Mate Here Is ur Answer....

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.Let's convert everything to no of moles first

1) 18g water→1 mole

2)18g Oxygen → 18/32 moles

3)18g Carbon dioxide→ 18/44

4)18 g methane → 18/16

clearly 18 g methane have maximum no of molecules as mor no of moles.

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. plZz Mark Me As Brainliest ✌

Answer 2

D) 18g of Methane $CH_{4}$ has the maximum number of atoms.

Calculating the molecular mass:

• Molecular mass is the mass of a single molecule calculated by the addition of atomic masses of all the atoms present in that particular molecule.

• First, we calculate the molecular mass of all the given compounds.

For $H_{2}O$, the molecular mass will be:

$\Rightarrow \ 2 \ \times \ atomic \ mass \ of \ hydrogen \ atom \ + \ atomic \ mass \ of \ oxygen$

$\Rightarrow 2 \times (1) + 16\\\\\Rightarrow 2 + 16\\\\\Rightarrow 18g$

Similarly, for $O_{2}, \ CO_{2}$ and $CH_{4}$, molecular masses will be 32g, 44g, and 16g respectively.

Calculating the number of moles (n):

• 'Moles' is the measurement unit to measure the amount of any chemical substance.
• According to the Internal Systems of Units, 1 mole of a chemical substance is equivalent to $6.023 \ \times \ 10^{23}$ (Avogadro's number) some chemical entity (can be atoms, molecules, and ions).

The formula for calculating the number of moles is:

$Number \ of \ moles \ = \ \frac{given \ mass}{molecular \ mass}$

Since the given mass of all the compounds is 18g, then, for $H_{2}O$,

$Number \ of \ moles \ = \ \frac{given \ mass}{molecular \ mass}\\\\ \Rightarrow \ \frac{18g}{18g}\\\\\Rightarrow \ Number \ of \ moles \ = 1 \ mol$

Similarly, for $O_{2}, \ CO_{2}$, and $CH_{4}$, the number of moles will be 0.5625, 0.4091, and 1.125 respectively.

Calculating the number of atoms:

We know, 1 mole is equivalent to the $6.023 \ \times \ 10^{23}$ (Avogadro's number).

Therefore, 1 mole  $H_{2}O$ has $6.023 \ \times \ 10^{23}$ atoms.

Similarly,

$\Rightarrow$ 0.5625 moles of $O_{2}$ has $0.5625 \ \times \ 6.023 \ \times \ 10^{23} \ = \ 3.388 \ \times \ 10^{23}$ atoms

$\Rightarrow$ 0.4091 moles of $CO_{2}$ has $0.4091 \ \times \ 6.023 \ \times \ 10^{23} \ = \ 2.464 \ \times \ 10^{23}$ atoms

$\Rightarrow$ 1.125 moles of $CH_{4}$ has $1.125 \ \times \ 6.023 \ \times \ 10^{23} \ = \ 6.776 \ \times \ 10^{23}$ atoms

Hence, 18g of methane $CH_{4}$ has $\ 6.776 \ \times \ 10^{23}$ atoms, which is the maximum.