A sample of sodium carbonate contains sodium also. 1.5 g of the sample is dissolved in water and volume raised to 250 mL. 25 mL of this solution requires 20 mL of `(N)/(10)H_(2)SO_(4)` solution for neutralisation. Calculate the percentage of sodium carbonate in the sample.
Text Solution
Verified by Experts
Only `Na_(2)CO_(3)` will react with `H_(2)SO_(4)`. Applying `underset((Na_(2)CO_(3)))(N_(1)V_(1)=underset((H_(2)SO_(4)))(N_(2)V_(2))` `N_(1)xx25=20xx(1)/(10)` `N_(1)=(20)/(25xx10)=0.08` Eq. mass of `Na_(2)CO_(3)=("Mol.mass")/(2)=(106)/(2)=53` Mass of `Na_(2)CO_(3)` present in 250 mL 0.08 N solution `(NxxExxV)/(1000)=(0.08xx53xx250)/(1000)=1.06g` Percentage of `Na_(2)CO_(3)` in the mixture `=(1.06)/(1.50)xx100=70.67`
Topper's Solved these Questions
VOLUMETRIC ANALYSIS
OP TANDON|Exercise Example 12|2 Videos
VOLUMETRIC ANALYSIS
OP TANDON|Exercise Example 13|2 Videos
VOLUMETRIC ANALYSIS
OP TANDON|Exercise Example 10|2 Videos
THE COLLOIDAL STATE
OP TANDON|Exercise Self Assessment|21 Videos
Similar Questions
Explore conceptually related problems
In a sample of sodium carbonate some sodium sulphate is also mixed. 1.25 g of this sample is dissolved and the volume made up to 250 mL. 25 mL of this solution neutralises 20 mL of (N)/(10) sulphuric acid. Calculate the percentage of sodium carbonate in the sample.
0.5 g of an oxalate was dissolved in water and the solution made to 100 mL. On titration 10 mL of this solution required 15 mL of (N)/(20)KMnO_(4) . Calculate the percentage of oxalate in the sample .
10.875 g of a mixture of NaCl and Na_(2)CO_(3) was dissolved in water and the volume made up to 250 mL, 20 mL of this solution required 75.5 mL of (N)/(10) H_(2)SO_(4) . Find out the percentage composition of the mixture.
4.0 g of a mixture of Nacl and Na_(2) CO_(3) was dissolved in water and volume made up to 250 mL . 25 mL of this solution required 50 mL of N//10 HCl for complete neutralisation. Calculate the percentage composition of the original mixture.
