Question

Problem 4: A waste water sample contains 2.092 mmol/L NaCl. Calculate the concentration of

  NaCl in mg/L. What are the mmol/L concentrations of Na and Cl. Na = 23 and

  Cl = 35.5.

Problem 5: Following analysis passes the acceptability of analysis (satisfies the

  electroneutrality). Find the concentration of Ca2+ mg/L.

Na+ = 12.8 mg/L ​​Cl- = 10.4 mg/L

Ca2+ = ??? ​​​HCO3 = 195.2 mg/L

Mg2+ = 15.5 mg/L ​​SO42- = 33.1 mg/L

K+ = 7.8 mg/L ​​​

Problem 6: Find the ionic strength of a 4 mM of CaCl2. What are the activities of Ca2+ and Cl-.

Answer 1

Given that;

2.092 mmol/L NaCl

1.0 mmol = 0.001 mol

0.001 mol/ 1.0 mmol * 2.092 mmol=2.092*10^-3 mol NaCl

Molar mass of NaCl – 58.5 g/mol

Now calculate the mass of NaCl in g:

2.092*10^-3 mol NaCl * 58.5 g/mol

= 0.122 g

Thus 0.122 g NaCl is present in 1.00 L.

1.0 g= 1000 mg

0.122 g * 1000 mg/1.0 g = 122 mg

Hence the concentration of   NaCl in mg/L is 122 mg /L.

What are the mmol/L concentrations of Na and Cl

One mole NaCl gives 1 mol of Na+ and 1 mol Cl-

NaCl -à Na+ +Cl-

Thus 2.092 mmol/L NaCl gives 2.092 mmol/L Na + and 2.092 mmol/L Cl-.