Drinking water was found to be served real call morning and chloroform supposed to be credited in nature nature the revolt of contamination with 15 PPM by mass
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1 ppm is equivalent to 1 part out of 1 million (106) parts.
∴ Mass percent of 15 ppm chloroform in water
(ii) molality (M) = no of moles of solute/mass of solvent in g *1000
Therefore mass of chloroform= 12 + 1+3(35.5) = 119.5 g/mol
100 g of the sample contains 1.5 × 10–3 g of CHCl3.
⇒ 1000 g of the sample contains 1.5 × 10–2 g of CHCl3.
m = 1.5 x 10-3/119.5 * 1000 = 1.25x 10-4 m
I know your question is incomplete.
But I have studied this question in earlier classes.So on basis of that I have answered your question.
ashwanisingh2001:
You are have been asked to find to express it in percent by mass and molarity of chloroform in water
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