Question

A sample of drinking water was found to be severely contaminated with chloroform, $CHCl_{3}$, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.

Answer 1

Formula to calculate mass percentage in parts per million =$\frac { Mass\quad of\quad the\quad solute }{ Mass\quad of\quad the\quad solution } \quad \times \quad { 10 }^{ 6 }$

On substituting the values:$\frac { 15 }{ { 10 }^{ 6 } } \quad \times \quad 100\quad =\quad 1.5\quad \times \quad { 10 }^{ -3 }%$

i) Percent by mass: $1.5\quad \times \quad { 10 }^{ -3 }%$

Molality is the concentration of the solution by calculating the number of solute moles/kg of the solvent

So, $Molality\quad (M)\quad =\quad x\quad =\quad \frac { No.of\quad solute\quad moles }{ Solvent\quad expressed\quad in\quad kilogram }$

Mass of chloroform: 119.38 g/mol $[CH{ Cl }_{ 3 } = 1 carbon +1 hydrogen + 3 chlorine atoms = 12 + 1+ 106.5 = 119.3]$

Since, the mass percentage of $CH{ Cl }_{ 3 }$  in 100 g of water sample is $1.5\quad \times \quad { 10 }^{ -3 }\quad g;$ 1000 g of water sample will contain:

$\frac { 1.5\quad \times \quad { 10 }^{ -3 } }{ 119.5\quad \times \quad 1000 } \quad =\quad 1.25\quad \times \quad { 10 }^{ -4 }\quad m$

ii) Molality of chloroform: $1.25\quad \times \quad { 10 }^{ -4 }\quad m$

Answer 2

Answer:

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