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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
$$1.25 \times 10^{-4}$$ mB
$$2.5 \times 10^{-4}$$ mC
$$1.5 \times 10^{-3}$$ mD
$$1.25 \times 10^{-5}$$ 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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