Question

Sup People!

A sample of drinking water was found to be contaminated with chloroform CHCl3. The level of contamination was 15ppm(by mass).
i) Express this in percent by mass.
ii) Determine the molality of chloroform in the water sample.

Also define:-
ppm

<You have to explain your each step, cause I am super confused about this question @_@, Pretty pleasee ;_;>

Don't Spam for Goodness sake,
Please help me out :]


​

Answer 1

Answer ;(i) 15 ppm means 5 parts in million(106) parts.

∴ % by mass = 15/106 × 100 = 15×10-4 = 1.5×10-3 %

(ii) Molar mass of chloroform(CHCl3) = 12+1+(3×35.5) = 118.5 g mol-1

100g of the sample contain chloroform = 1.5×10-3g

∴ 1000 g(1 kg) of the sample will contain chloroform = 1.5×10-2 g

= 1.5×10-2/118.65 mole = 1.266×10-4 mole

∴ Molality = 1.266×10-4 m.

Explanation:

Answer 2

Answer:

Sup Sis!

(i) 15 ppm means 5 parts in million(106) parts.

∴ % by mass = 15/106 × 100 =

$15 \times \: 10 {}^{ - 4} = 1.5 \times \: 10 {}^{ - 3}$

(ii) Molar mass of chloroform(CHCl3) = 12+1+(3×35.5) = 118.5 g mol-1

100g of the sample contain chloroform =

$1.5 \times 10 {}^{ - 3}$

∴ 1000 g(1 kg) of the sample will contain chloroform =

$1.5 \times 10 {}^{ - 2}$

= 1.5×10-2/118.65 mole = 1.266×0-4 mole

∴ Molality =

$1.266 \times 10 {}^{ - 4} m.$