Question

What weight of calcium contains the same number of atoms as are present in 3.2 g of sulphur ?​

Answer 1

Given :

• given weight of sulphur = 3.2 g

To Find :

• The weight of calcium which contains same number of atoms as present in 3.2 g of sulphur

Knowledge required :

• Number of moles of a given substance is calculated by ,

$\\ \: {\boxed{\purple{\sf{n = \frac{given \:weight}{atomic \: weight} }}}}$

• Number of atoms of a given substance is given by ,

$\\ \: {\boxed{\sf{\purple{N = N_A \times n}}}}$

[ Where N is number of atoms , $\sf{N_A}$ is avogadro number and n is number of moles ]

Solution :

Given weight of sulphur = 3.2 g

Atomic weight of sulphur = 32 g/mol

Using the number of moles formulae . Number of moles of sulphur is ,

$\\ : \implies \sf \: n = \frac{3.2}{32}$

$\\ : \implies{\boxed{\mathfrak{\red{\: n = 0.1 \: mol}}}} \:$

Now calculating number of atoms in given amount of sulphur ,

Avogadro number = 6.023 × 10²³

$\\ : \implies \sf \:N = 6.023 \times {10}^{23} \times 0.1$

$\\ : \implies\boxed{\sf{\pink{N = 6.023 \times {10}^{22} }}}$

Let the weight of calcium required be x g

Atomic weight of calcium = 40 g/mol

$\\ : \implies \sf \: n = \frac{x}{40} \: ............(1)$

Number of atoms in calcium is ,

$\\ : \implies \sf \: N = \: 6.023 \times {10}^{23} \times \frac{x}{40}$

Here Number of atoms in x gram of calcium is equal to the number of atoms in 3.2 g of sulphur [given].

$\\ : \implies \sf \: 6.023 \times {10}^{22} = 6.023 \times {10}^{23} \times \frac{x}{40}$

$\\ : \implies \sf \: 1 = 10 \times \frac{x}{40}$

$\\ : \implies \sf \: 40 = x \times 10$

$\\ :\implies{\boxed{\pink{\mathfrak{ \: x = 4 \: g}}}} \bigstar$

Hence ,

• 4g of calcium contains same number of atoms as present in 3.2 g of Sulphur.

Answer 2

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