Answer :
1.006 g of Na2CO3 is left; which means that (2.714 -1.006) g of water are lost from the hydrated sodium carbonate
So, water present = (2.714 -1.006) = 1.708 g.
Molecular Wt. of water is =18 g
so moles = 1.708/18 moles = 0.095 moles.
Now, Na2CO3 has molecular wt. of = 23x2 + 12 +3x16 = 106 g,
so moles of Na2CO3 = 1.006/106 moles = 0.0095 moles
Ratio is = 0.0095 / 0.095 = 1 / 10 [ moles of Na2CO3 / moles of H2O ]
Ratio is saying on every 1 mole of Na2CO3 , there are 10 moles of H2O
so x is 10
[tex]Na_2CO_3.10H_2O[/tex]
So, water present = (2.714 -1.006) = 1.708 g.
Molecular Wt. of water is =18 g
so moles = 1.708/18 moles = 0.095 moles.
Now, Na2CO3 has molecular wt. of = 23x2 + 12 +3x16 = 106 g,
so moles of Na2CO3 = 1.006/106 moles = 0.0095 moles
Ratio is = 0.0095 / 0.095 = 1 / 10 [ moles of Na2CO3 / moles of H2O ]
Ratio is saying on every 1 mole of Na2CO3 , there are 10 moles of H2O
so x is 10
[tex]Na_2CO_3.10H_2O[/tex]