Ncert Exercises & Solutions

1.17 A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, 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.

(i) 1 ppm is equivalent to 1 part out of 1 million (10⁶) parts.
Mass percent of 15 ppm chloroform in water $=$ $15 / 10^{6} x 100 = 1.5 x 10 - 3 %$

(ii) 100 g of the sample contains 1.5 × $10^{- 3}$ g of ${CHCl}_{3}$ .
⇒ 1000 g of the sample contains 1.5 × $10^{- 2}$ g of ${CHCl}_{3}$ .
$∴$ Molality of chloroform in water

$= \frac{}{1.5 \times 10 - 2 g / M o l a r m a s s o f C H C l 3}$

Molar mass of CHCl3 = 12.00 + 1.00 + 3(35.5)
= 119.5 g ${mol}^{- 1}$
Molality of chloroform in water = 0.0125 × $10^{- 2}$ m
= 1.25 × $10^{- 4}$ m

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