Question

0.715 g of Na2CO3 x H2O required 20 ml of seminormal hydrochloric acid solution
for complete reaction. Find the value of x.​

Answer 1

Answer:

Explanation:

equivalent of 1 = equivalent of 2

0.715 / 106+18x * 1*1000  =  10

1430/106+18x = 10

1430=1060+180x

x=2.05

Answer 2

The number of moles of x in the formula is 2

Given:

0.715g of Na₂CO₃xH₂O

Volume of seminormal HCl = 20ml

To find:

The value of the number of water molecules present in the given molecule

Solution:

For titration,

M₁V₁n₁ = M₂V₂n₂

We know that the semi-normal equivalent is 1/2

Therefore the equivalent of HCl is given by,

= 0.02 x 0.5

= 0.01 equivalent HCl

Equivalent weight of 0.715g of Na₂CO₃xH₂O = 0.715/0.01 = 71.5 equivalent weight of Na₂CO₃xH₂O

The equivalent weight of Na₂CO₃ = 106/2 = 52 equivalent weight of   Na₂CO₃

Molar weight of Na₂CO₃xH₂O = Molar weight of Na₂CO₃+MW xH₂O =143g

Molar weight of H₂O  = Molar weight of Na₂CO₃xH₂O -Molar weight of Na₂CO₃

                                     = 143- 106

Molar weight of H₂O   = 37 g of water

No of moles of H₂O(x) = 37 / 18

                                     = 2.06

Therefore,
x = 2

The number of moles of water molecule in 0.715g of Na₂CO₃xH₂O is 2. Hence the formula will be Na₂CO₃2H₂O

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