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any one help me in this prb
0.7g of Na^2CO^3.XH^2O is dissolved in 100mlH^20,20ml of which required 19.8 ml of 0.1N HCL.the value of x is
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7 years ago

23 Points
```										Dear Jauneet,
We know that the hydrolysis of anhydrate sodium carbonate give 2 moles of sodium hydroxide.
So there is already x amount of water present in the sample so amount of water it contributes is x*18*density of water = 18x
We know that molarity= no of moles per litre of solution so the molecular weight of the given compound is (106+18x)gm/ mol and the total amount of water added is (100+18x) and we also know that basicity of the given compound is 2 two as it gives two moles of NaOH on hydrolysis. we also have the formula N1 * V1 =N2 * V2
=> 20*2*(0.7/(106+18x)) * (100/(100+18x)) = 0.1* 19.8
on solving this equatiopn we get x= 0.886 moles

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```
7 years ago
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