Question

Calculate the amount of reagent and water required to make the following solutions: 1.) 650 mL of 25 mM MgCl₂ 2.) 400 mL of 5% Na₂HCO₃ 3.) 25 mL of 2.5 M NaCl 4.) 1.5 L of 45 uM CaCl₂

Answer 1

Molarity = no. of moles / volume of solution in litres

mM denotes milli molar or 10^-3 M

25 mM = 25 x 10^-3 M

water required would be 650 mL = 0.65L

putting the values in equation

25 x 10^-3 M = moes of solute(n)/0.65 L

n = 0.01625 moles

Molar mass of MgCl₂ = 95.211 g/mol

Mass = moles x molar mass = 0.01625 mol x 95.211 g/mol = 1.547 g

or 1547 mg

2)

% = mass of solute / mass of solution x 100

5 = mass of solute(m) /{400g + m}x100 density of water =1g/mL and 400 mL = 400g

2000g + 5m = 100m

95m = 2000g

m = 2000g/95

m = 21.0526 g

3)

although in water NaCl will not remain as NaCl but as ions Na+ and Cl- and precisely it should be called as formality

but leaving that fact we can calculate

similarly as 1)

2.5 M = m/ 0.025 L

m = 0.0625 moles

Molar mass of NaCl = 58.44 g/mol

0.0625 moles = 0.0625 mol x 58.44 g/mol = 3.652 g

4)

45 uM = 45 x 10^-6 M = moles of solute(n)/1.5 L

n = 67.5 x 10^-6 moles

or = 67.5 uM

molar mass of CaCl₂ = 110.98 g/mol

mass of CaCl₂ = moles x molar mass = 67.5 uM x 110.95 g/mol = 7489.12 ug or 0.0749 g