What will be the molality of chloroform in the water sample which contains 15 ppm chloroform by mass

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Matthew M.

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11 months ago

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calculate the molality. (d) 0.372 g of histamine, C5H9N, in 125 g of chloroform, CHCl3

What will be the molality of chloroform in the water sample which contains 15 ppm chloroform by mass?

• A

  $$1.25 \times 10^{-4}$$ m

• B

  $$2.5 \times 10^{-4}$$ m

• C

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

• D

  $$1.25 \times 10^{-5}$$ m

Concentration of chloroform=15 ppm

$$\therefore$$,10^6 g of solution contains 15 g chloroform 

thus 1000g of solution contains $$\cfrac{15}{10^6} \times 1000=15 \times 10^{-3}g$$ chloroform

Molality=$$\cfrac{number\ of\ moles\ of\ solute}{mass\ of\ solvent(kg)}$$

and number of moles=$$\dfrac{mass}{molar\ mass}$$

for Chloroform $$CHCl_3$$, mass$$=15 \times 10^{-3}g$$, molar mass=$$12+1+3 \times 35.5=119.5 g/mol$$

and water, mass=1000g=1 kg

thus Molality=$$\cfrac{15 \times 10^{-3}}{119.5 \times 1}=1.25 \times 10^{-4}m$$

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

Therefore, mass percent of 15 ppm chloroform in water

= (15/106) x100 =1.5 x 10-3 %

ii) 100 g of the sample contains 1.5 x 10-3 g of CHCl3.

1000 g of the sample contains 1.5 x 10-2 g of CHCl3

Thus, Molality of chloroform in water

= 1.5 x 10-2 g/Molar mass of CHCl3

Molar mass of CHCl3 =12+1+3(35.5)

=119.5 g mol-1

Therefore, molality of chloroform in water =0.0125 x 10-2 m

=1.25 x 10-4 m

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