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# What will be the molality of chloroform in the water sample which contains 15 ppm chloroform by mass?A.1.25 x ${ 10 }^{ -4 }$ mB.2.5 x ${ 10 }^{ -4 }$ mC.1.5 x ${ 10 }^{ -3 }$ mD.1.25 x ${ 10 }^{ -5 }$ m

Last updated date: 20th Jun 2024
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Hint: To answer this question first try to find the weight of chloroform present in water then using the density of water, we can easily convert this value to the molality of water.

Here we should know that X ppm means in solution means X parts in the one million ( ${ 10 }^{ 6 }$) parts.
Since we are given 15 ppm chloroform,
Therefore,
% by mass of chloroform = $\dfrac { 15 }{ { 10 }^{ 6 } } \times 100$ = 1.5 x ${ 10 }^{ -3 }$ %
Now, we know the formula of molality in terms of % by mass is given by,
$Molality(m)=\dfrac { Percent\quad by\quad mass\times density\quad of\quad solution\times 10 }{ Molar\quad mass\quad of\quad solute }$
Where,
% by mass = 1.5 x ${ 10 }^{ -3 }$ %
density of solution = 1g/${ cm }^{ 3 }$
Molar mass of solute (${ CHCl }_{ 3 }$) = 119 g/mol
Now, insert these values in the equation,
$Molality(m)\quad =\quad \dfrac { 1.5\times { 10 }^{ -3 }\times 1\times 10 }{ 119 }$
$Molality(m)\quad =\quad \dfrac { 1.5\times { 10 }^{ -3 } }{ 119 }$
$Molality(m)\quad =\quad 1.25\times { 10 }^{ -4 }\quad m$

Therefore, the correct answer to this question is option A, $1.25\times { 10 }^{ -4 }\quad m$.

Note:
We should know that Concentrations expressed in molality are used when studying the properties of solutions related to vapor pressure and temperature changes. It is used because its value does not change with changes in temperature.