# A sample of `Na_2SO4` is contaninated with some `NaCl` . The following procedure was used to determine the percentage of `NaCl` in the sample. 1.000 g of the sample was dissolved in water in a 250...

A sample of `Na_2SO4` is contaninated with some `NaCl` . The following procedure was used to determine the percentage of `NaCl` in the sample.

1.000 g of the sample was dissolved in water in a 250 cm^3 volumetric flask and diluted upto the mark to form the solution A.

The following solution (1)-(5) wer prepared by diluting a `4.0 xx 10^-4 mol. dm^-3` `Cl^-` ion solution.

1 2 3 4 5

4.0 * mol.dm-3 Cl^- ion solution/cm^3 1.00 2.00 3.00 4.00 5.00

Distilled water/cm^3 9.00 8.00 7.00 6.00 5.00

Into each of the solution (1)-(5).`1.00 cm^3` of dill.` HNO_3` and `1.00 cm^3` `AgNO_3` were added. Similarly into `10.00 cm^3` of solution A, `1.00 cm^3` dill `HNO_3` and `1.00 cm^3` `AgNO_3` were added.

It was observed that the turbidity produce by solution A was equal to that of solution (3). Calculate the mass percentage of NaCl in the sample of `Na_2SO_4` .(Na=23.0,Cl=35.5)

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Concentration of `Cl^- ` in solution No.3

= `(4.0 xx 10^-4 xx 3)/(10) ` `mol dm^-3`

=`1.2 xx 10^-4` `mol dm^-3`

`` Molar mass of `NaoH` = `(23.0+35.5 )` `g.mol^-1` `=58.5 g.mol^-1`

The mass of `NaCl` `1.0 dm^3` of solution `=1.2 xx 10^-4 xx 58.5 g`

The mass of sample in `1.0 dm^-3 = 4.00 g`

Percentage of `NaCl` in the sample `=(1.2 xx 10^-4 xx 58.5 g xx 100)/(4.00 g)` = `=0.1755`