Question

Calculate the number of molecules present in 12.3 g MgSO4 -7H2O. Calculate the mass of Na2CO3 which will

have molecules equal to those present in 12.3g of MgSO4 .7H2O.

Answer 1

In First case,
Mass of the MgSO₄.7H₂O = 12.3 grams.
Molar Mass of the MgSO₄.7H₂O = 246 g/mol.

∴ Using the Formula,
 No. of Molecules = ( Mass/Molar Mass) × Avogadro's number. 
  = (12.3/246) × 6.022 × 10²³
  = 0.05 × 6.022 × 10²³
  = 0.3011 × 10²³
  = 3.01 × 10²²

Hence, the number of the molecules in the MgSO₄.7H₂O is 3.01 × 10²² molecules. 

In Second Case,
Molar Mass of the Na₂CO₃ = 106 g/mol.

∵ No. of molecules of the MgSO₄.7H₂O is equals to the number of the molecules of Na₂CO₃. 
∴ 3.01 × 10²² =  (Mass/106 ) × 6.022 × 10²³
∴ Mass = (106 × 3.01)/60.22
   = 319.06/60.22
   = 5.29
  ≈ 5.3 grams.

Hence, the mass of the Na₂CO₃ is 5.3 grams.


Hope it helps.

Answer 2

Mass of the MgSO4 -7H2O is 12.3 grams. 

∵ 6.022 × 10²³ molecules of the  MgSO4 -7H2O is present in the 1 mole of it. 
∴ 1 mole of the  MgSO4 -7H2O contains 6.022 × 10²³ molecules.
∴ 246 g of the  MgSO4 -7H2O contains  6.022 × 10²³ molecules. 

Thus, 12.3 g of the  MgSO4 -7H2O contains 12.3 × 6.022 × 10²³/246 molecules.
 = 0.3011 × 10²³ molecules. 


From the Question, number of molecules of the Na₂CO₃ is 0.3011 × 10²³ molecules. (same as  MgSO4 -7H2O) 

∵ 1 mole of the Na₂CO₃ contains 6.022 × 10²³ molecules.
∴ 106 g of the Na₂CO₃ contains 6.022 × 10²³ molecules.
or 1 molecules of the Na₂CO₃ weighs 106/(6.022 × 10²³)
∴ 0.3011 × 10²³ molec. of the Na₂CO₃ weighs 106/(6.022 × 10²³) × 0.3011 × 10²³         =  (106 × 0.3011)/6.022 = 5.3 grams.


Thus, the mass of the Sodium carbonate is 5.3 grams.


I hope it will helps you.
Thanks for asking the wonderful Question.