Question

A 15.2% (m/v) solution of sodium chloride (NaCl) is used to prepare 25.0 L of a physiological saline solution with a concentration of 0.154 M NaCl. What volume of the 15.2% (m/V) solution of sodium chloride would be required to make 25.0 L of the physiological saline solution ( 0.154 M NaCl) ?

Answer 1

Ans. Calculating molarity of stock solution

15.2% (w/v) NaCl solution means that 100 mL of the solution contains 15.2 g NaCl.

Moles of NaCl in 100 mL solution = Mass of NaCl / Molar Mass of NaCl

                                                = 15.2 g / 58.44 g mol^-1

                                                = 0.26 moles

Molarity of 15.2 % w/v solution = Moles of NaCl / Volume of solution in Liters

                                                = 0.26009 moles / 0.100 L

                                                = 2.6009 M = 2.60 M (approx.)

Now,

Using, M1V1 = M2V2   -- equation 1

where

M1= Molarity of initial solution 1, V1= volume of initial solution 1      ;( stock solution)

M2= Molarity of final solution 2, V2= volume of final solution 2 ;(diluted solution i.e. Normal saline)

Given,

            M1 = 2.60 M

            V1 =?

            M2 = 0.154 M

            V2 = 25.0 L

Putting the values in equation 1-

            2.60 M x V1 = 0.154 M x 25.0 L

            Or, V1 = (0.154 M x 25.0 L) / 2.60 M = 1.481 L

Thus, the volume of stock solution required = 1.481 L