Lab Report - What volume of concentrated 12N HCl is required to prepare 250 cm3 of 5N HCl solution?

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       Chemistry

       

What volume of concentrated 12N HCl is required to prepare 250 cm3 of 5N HCl solution?

   

         

       Lab Report

        

 

Experimental Design

Aspect 1

Variables

Responding Variables –

  1. Normality of HCl solution – 12N

Controlled Variables –

  1. Volume of 5N HCl solution to be prepared – 250 cm3
  2. Volume of 12N solution required – 104.16 cm3 

Aspect 2

Hypothesis

M1*V1 = M2*V2 

So, V1 = (M2*V2) / M1

V1 = (5 * 250) / 12

Thus, V1 = 104.16 cm3

Therefore, 104.16 cm3 of 12N HCl is required to prepare 250 cm3 of 5N HCl solution.

Aspect 3

Controlling Variables

  1. Use a parallax card while measuring the amount of concentrated HCl in order to avoid any sort of error.
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Aspect 4

Materials and apparatus required

  1. Measuring Cylinder (250 cm3) x1
  2. Measuring flask (250 cm3) x 1
  3. Beaker (500 cm3) x 1
  4. Glass rod x 1
  5. Safety Glasses x 1
  6. Wash bottle (500 cm3) x 1
  7. A pair of gloves x 1

Chemicals Required

  • Distilled Water (500 cm3)
  • 12N HCl (104.2 cm3)

Aspect 5

Procedure

  1. Accurately measure out 104.2cm3 of 12N concentrated HCl in a measuring cylinder.
  2. Take approximately around 100-120 ...

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