A 118-mg sample of Na2CO3 is dissolved in H2O to give 2.30L of solution.
The molar mass of Na2CO3=23x2+12+16x3=106 g/mol
So the molar of Na2CO3=118x10-3/106=1.11x10-3 mol
1mol Na2CO3 will produce 2mol Na+ in water, So 1.11x10-3 mol Na2CO3 will produce 2.22x10-3 mol Na+ in water.
2.22x10-3 mol Na+=2.22x10-3x23 g Na+=5.11x10-2g Na+
Assuming the density of the solution is the same as the density of water,
2.30L solution=2.30x1000g/L=2.30x10 3 g solution
Therefore the concentration of Na+= 5.11x10-2/2.30x10 3=2.22x10-5=22.2x10-6=22.2 ppm
The molar of HClO4=0.803x40x10-3=32.12x10-3 mol
So the final molarity of HClO4=32.12x10-3 mol/660.00x10-3 L=0.0487 mol/L=0.0487 M
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