assume that the density of all solutions are 1.000g/ml 1. Calculate the molarity of calcium in 1.9g of calcium chloride diluted in 100 ml of Di water. 2 Calculate the concentration of both calcium and chloride lons in problem 1 in units of mg/mL, ug/L, mg/L and ug/mL. 3. Calculate the concentration of both calcium and chloride ion in problem 1 in units of ppm and ppb. You may assume that the density of the solution is 1.0 g/ml 4. You have been provided 100 ml of a 1000 ug/ml barium standard. What volume of this standard must be diluted to a final volume of 50 ml using DI water to produce a 30 ug/mL standard

1. 0.1712M

2. 6.86mg/mL Ca, 12.14mg/mL Cl, 6860000ug/L Ca, 12140000ug/L Cl, 6860mg/L Ca, 12140mg/L Cl, 6860ug/mL Ca, 12140ug/mL Cl.

3. 6860ppm Ca and 12140ppm of Cl.

4. 1.5mL of the 1000ug/mL barium standatd must be taken.

Explanation:

1. Molarity is defined as the amount of moles of solute (Calcium chloride) present in 1L of solution.

The moles of CaCl₂ are:

1.9g CaCl₂ * (1mol / 110.98g) = 0.01712 moles

In 100mL = 0.10L:

0.01712mol / 0.10L = 0.1712M

2. The masses of Calcium and Chloride ions are:

1.9g * (40.078g Ca / 110.98g) = 0.686g Ca

And:

1.9g - 0.686g Ca = 1.214g Cl

mg/mL:

686mg Ca / 100mL = 6.86mg/mL Ca

1214mg Cl / 100mL = 12.14mg/mL Cl

ug/L:

686000ug / 0.1L = 6860000ug/L Ca

1214000ug/ 0.1L = 12140000ug/L Cl

mg/L:

686mg Ca / 0.1L = 6860mg/L Ca

1214mg Cl / 0.1L = 12140mg/L Cl

ug/mL:

686000ug Ca / 100mL = 6860ug/mL Ca

1214000ug Cl / 100mL = 12140ug/mL Cl

3. ppm are defined as mg/L, the ppm of Ca are 6860ppm Ca and 12140ppm of Cl

4. The solution must be diluted from 1000ug/mL to 30ug/mL, that is a dilution of:

1000ug/mL / 30ug/mL:

33.33 times must be diluted the solution.

As final volume of the diluted solution must be 50mL, the volume of the standard needed is:

50mL / 33.33 times = 1.5mL of the 1000ug/mL barium standatd must be taken