Question

Calculate the number of sodium ions that are present in 212g of sodium carbonate? Answer is 2.4092×10²⁴

Answer 1

Answer: 4 x 6.02 x 10^23

Step-by-step explanation:

Relative atomic masses

Na = 23

C = 12

O = 16

Molar mass of Na2CO3 = 23x2 + 12 + 16x3 = 106g/mol

Moles of Na2CO3 = Given mass/Molar mass

                              = 212/106

                               = 2 moles

One mole of Na2CO3 has two moles of sodium ions

∴ Moles of sodium ions = 2 x 2 = 4  

Number of sodium ions = Number of moles x Avogadro’s number

= 4 x 6.02 x 10^23

Answer 2

Answer:

24.088 × 1023 ions

Step-by-step explanation:

Mass of sodium carbonate = 212g 

Sodium carbonate is Na2CO3 

First, we need to calculate the number of moles of sodium carbonate we have in a 212g sample.

To calculate this, we will ind the molar mass of sodium carbonate (Na2CO3 ):  ⇒ 2 × Molar mass of sodium + Molar mass of carbon + 3×molar mass of oxygen 

⇒ 2 × 23 + 12 + 3 × 16  ⇒ 46 + 12 + 48  ⇒ 106g/mol 

Thus, the molar mass of Na2CO3 is 106g/mol.  Now we will calculate total number of moles by applying the formula given:

Number of moles = Mass of sodiumcarbonate Molar mass of sodiumcarbonate Mass of sodium carbonate

Molarmassofsodiumcarbonate

 Number of moles = 212g106g/mol212g106g/mol

⇒ 2 mol 

Now, we know that every mole of Na2CO3 have 2 moles of Na+ ions

.Hence, total moles of Na2CO3 is 4 moles Number of ions present = 6.022 × 1023 × 4mol

= 24.088 × 1023 ions