Question

Na_(2)CO_(3) have water of crystallisation and its formula is Na_(2)CO_(3)*xH_(2)O .If 2.32g of Na_(2)CO_(3)*xH_(2)O is heated then it loose 1.26g of water the value of x is Assume all water of crystalline are removed when Na_(2)CO_(3)*xH_(2)O is heated​

Answer 1

Answer:

Explanation:

Mass of hydrous=2.32g

Mass of anhydrous=1.26g

Na2CO3.xH2O=106+18X

2.32/1.26=106+18X/106

1.26(106+18X)=246

106+18X=195

18X=89

X=5

Therefore Na2CO3.5H2O

Answer 2

CORRECT QUESTION

Na₂CO₃ have water of crystallisation and its formula is Na₂CO₃.xH₂O .If 2.32g of Na₂CO₃.xH₂O is heated then it loose 1.26g of water the value of x is ? Assume all water of crystalline are removed when is heated.

ANSWER

The value of x is 7.

GIVEN

Mass of Na₂CO₃.xH₂O = 2.32 grams

Mass of water of crystallisation lost = 1.26 grams

TO FIND

Value of x

SOLUTION

We can simply solve the above problem as follows;

On heating Na₂CO₃.xH₂O following reaction takes place;

Na₂CO₃.xH₂O -------------> Na₂CO₃ + xH₂O

Mass of Na₂CO₃.xH₂O = 3.32 grams

According to law of conservation of mass

2.32grams = Mass of Na₂CO₃ + Mass of xH₂O

Mass of anhydrous Sodium Carbonate = 2.32 - 1.26 = 1.06

Molar mass of Na₂CO₃ = 106 g/mol

Moles of Na₂CO₃ formed = 1.06/106 = 0.01 mole

Now,

Molar mass of H₂O = 18 g/mol

Mass of H₂O formed = 1.26 g

Moles of H₂O formed = 0.07 mol

When 0.01 moles of Na₂CO₃ is present 0.07 moles are present.

When 1 mole of Na₂CO₃ is present, Moles of H₂O = 7 moles.

Hence, The value of x is 7.

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