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0.7 grams of Na(2)CO(3).xH(2)O were diss...

0.7 grams of `Na_(2)CO_(3).xH_(2)O` were dissolved in water and the volume was made to 100 mL, 20 mL of this solution required 19.8 mL of N/10 HCl for complete neutralization. The value of x is Report your answer by rounding it upto nearest whole number.

Text Solution

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Number of equivalents of `HCl=(19.8xx0.1)/(1000)`
`=Number of equivalent of "Na_(2)CO_(3)" in 20 ml"`
`"Number of equivalents of "Na_(2)CO_(3)" in 100 ml"`
`=(19.8xx0.1)/(1000)xx(100)/(20)`
`because" "(19.8xx0.1)/(1000)xx(100)/(20)" equivalents = 0.7 g"`
`"1 equivalent "=(07xx1000 xx20)/(19.58xx0.1xx100)=70.7`
Thus equivalent weight = 70.7
Molecular weight = 141.4
Now `Na_(2)CO_(3)xH_(2)O=141.4`
`23xx2+60+18x=141.4`
`18x=35.4`
`x=1.97 =2`
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