Question

The atomic mass of He, C, Fe and Ag are 4 u, 12 u, 56 u and 108 u respectively. Which of the following has maximum number of atoms? (A) 12 g of He .(B) 12 g of C (C) 84 g of Fe (D )216 g of Ag ​

Answer 1

12g of He have maximum no. of atoms.

here 3 moles of He are present,

1mole of C,

1.5mole of Fe,

and 2mole of Ag.

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Answer 2

The maximum number of atoms are present in 12 grams of helium that are $1.8066\times 10^{24}$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

According to mole concept:

1 mole of an element contains $6.022\times 10^{23}$ number of atoms

For the given options:

• For A: 12 g of Helium

Given mass of helium = 12 g

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$\text{Moles of helium}=\frac{12g}{4g/mol}=3mol$

Number of atoms = $(3\times 6.022\times 10^{23})=1.8066\times 10^{24}$

• For B: 12 g of Carbon

Given mass of carbon = 12 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon}=\frac{12g}{12g/mol}=1mol$

Number of atoms = $(1\times 6.022\times 10^{23})=6.022\times 10^{23}$

• For C: 84 g of Iron

Given mass of iron = 84 g

Molar mass of iron = 56 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron}=\frac{84g}{56g/mol}=1.5mol$

Number of atoms = $(1.5\times 6.022\times 10^{23})=9.033\times 10^{23}$

• For D: 216 g of Silver

Given mass of silver = 216 g

Molar mass of silver = 108 g/mol

Putting values in equation 1, we get:

$\text{Moles of silver}=\frac{216g}{108g/mol}=2mol$

Number of atoms = $(2\times 6.022\times 10^{23})=1.2044\times 10^{24}$

Learn more about mole concept:

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