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A sample of `Na_2SO4` is contaninated with some `NaCl` . The following procedure was...

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krokdoor | eNotes Newbie

Posted October 21, 2013 at 2:27 PM via web

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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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nikolao | High School Teacher | (Level 1) Adjunct Educator

Posted October 21, 2013 at 2:31 PM (Answer #1)

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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`

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