Question

Calculate the number of molecules in 50 g of CaCo3

Answer 1

Answer :-

3.011×10²³ molecules

Explanation :-

In CaCO₃ , three elements are present i.e. Calcium, Carbon and Oxygen.

• Atomic mass of Calcium (Ca) = 40u

• Atomic mass of Carbon (C) = 12u

• Atomic mass of Oxygen (O) = 16u

Hence, molecular mass of CaCO₃ :-

= 40 + 12 + 16×3

= 52 + 48

= 100u

∵ Molecular mass of CaCO₃ is 100u.

∴ Molar mass of CaCO₃ is 100g/mol .

Number of moles in 100g CaCO₃ :-

= Given Mass/Molar mass

= 50/100

= 0.5 mole

• Avogadro Number = 6.022×10²³

Number of molecules :-

= No of moles × Avogadro Number

= 0.5×6.022×10²³

= 3.011×10²³ molecules

Thus, 3.011×10²³ molecules are there in 50g of CaCO₃ .

Answer 2

$\large{\boxed{\boxed{\sf Let's \: Understand \: Question \: F1^{st}}}}$

This is an simple question in which we have given 50g of Calcium Carbonate (CaCO₃) and have to find no. of molecules in it.

$\large{\boxed{\boxed{\sf How \: to \: do \: it?}}}$

Here F1st we simply calculate molecular mass of CaCO₃ it will be 100g we, know 100g of CaCO₃ contains 6.022×10²³ molecules [Avagadro no.] . So, for calculating no. of molecules in 50g of CaCO₃ we divide it by 100 and then multiply it with 50.

$\huge{\underline{\boxed{\sf AnSwer}}}$

___________________________

Given:-

• 50g of CaCO₃

Find:-

• No. of molecules.

Solution:-

Molecular Mass of CaCO₃ = 40 + 12 + 3×16

= 40 + 12 + 48

= 100g

$\sf \because 100g \: CaCO_3 \: contains \: 6.022 \times {10}^{23} molecules.$

$\sf \implies 1g \: CaCO_3 \: contains = \dfrac{6.022 \times {10}^{23}}{100}molecules$

$\sf \implies 1g \: CaCO_3 \: contains = 0.06022 \times {10}^{23}molecules$

$\sf \implies 50g \: CaCO_3 \: contains = 0.06022 \times {10}^{23} \times 50 \: molecules$

$\sf \implies 50g \: CaCO_3 \: contains = 0.06022 \times 50 \times {10}^{23} \: molecules$

$\sf \implies 50g \: CaCO_3 \: contains = 3.011\times {10}^{23} \: molecules$

$\underline{\boxed{ \sf\therefore 50g \: of \: CaCO_3 \: contains \: 3.011\times 10^{23} \: molecules}}$