Question

A.) Given 1.2 L of a 2.2 M sodium sulfate solution, calculate the volume in Liters you would need to add to obtain a 0.5 M solution of sodium sulfate. Remember you start with 1.2 L.
  
B.) What volume (in Litres) of 0.100 M sodium chloride solution can be prepared from 117 g of the salt?

Answer 1

A) For dilution purpose we use formula M1V1 = M2V2

M1 = initial concentration = 2.2

M2 = final concentration = 0.5 M

V1= 1.2 L

we find final volume V2

2.2 x 1.2 = 0.5 x V2

V2 = 5.28 L

B) NaCl mass = 117 g

NaCl moles = mass / Molar mass of NaCl

             = 117 / 58.44 = 2

We have formula

Molarity = moles of solute / volume fo solution

where Molarity is given 0.1 M and moles of NaCl = 2 , we find volume of solution

Hence 0.1   = 2 / volume of solution

volume = 20 L