Question

How many water molecules are present in 12,3 g MgSO4.7H2O?

Answer 1

Explanation:

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.