Question

A sample of drinking water was found to be contaminated with chloroform. the level of contamination was 15 ppm. the molality of chloroform in the water sample is then value of the 3 digit integer abc is :

Answer 1

(i) 1 ppm is equivalent to 1 part out of 1 million (106) parts.

∴ Mass percent of 15 ppm chloroform in water

 

 

(ii) molality (M) = no of moles of solute/mass of solvent in g *1000

 

Therefore mass of chloroform= 12 + 1+3(35.5) = 119.5 g/mol

 

100 g of the sample contains 1.5 × 10–3 g of CHCl3.

 

⇒ 1000 g of the sample contains 1.5 × 10–2 g of CHCl3.

 

 m = 1.5 x 10-3/119.5 * 1000 = 1.25x 10-4 m

Answer 2

Hey !!

Let the mass of solution be 10⁶ g.

Mass of solute, chloroform = 15 g

(1) % by mass of chloroform = Mass of chloroform / Mass of solution × 100

             = 15 g / 10⁶ g  × 100

             = 15 × 10⁻⁴ %

(2) Mass of solvent, water = 10⁶ g - 15 g ≈ 10⁶ ≈ 10³ kg

 Number of moles of chloroform, CHCl₃ = Mass of chloroform / Molar mass

                = 15 g / 119.5 g mol⁻¹  

                = 0.126 mol

         Molality of solution = Number of moles of chloroform / Mass of water (in kg) = 0.126 mol / 10³ kg

FINAL RESULT = 1.26 × 10⁻⁴  mol kg⁻¹