A sample of drinking water was found to be severely contaminated with chloroform

Correct Answer - `~15xx10^(-4) g ,1.25 xx10^(-4) m` (i) 1 PPm is equivalent to part out of million `(10^(6))` parts ` therefore ` mass percent of 15 ppm chloroform in water `=(15)/(10^(6))xx100` `~- 1.5 xx10^(-3)%` (ii) 100g of the sammple containts `1.5 xx10^(-3) g of CHCl_(3)`. `implies` 1000 g of the sample contains `1.5xx10^(-2) g ofCHCl_(3)` `therefore ` Molality of chloofromn in water `=(1.5 xx10^(-2) g)/("Molar mass of CHCl_(3))` Molar mass of `CHCl_(3) =12.00 +1.00 + 3(35.5)` `=119.5 g mol ^(-1)` ` therefore` Molality of chlorofrom in water `=0.0125xx10^(-2)` m `=1.25xx10^(-4) m`

This question hasn't been solved yet Ask an expert Ask an expert Ask an expert done loading Question:A sample of drinking water was found to be severely contaminated with chlorofor CHCl _(3), supposed to be carcinogenic in nature. The level of contamination was ppen [by mass ). (i) Express this in percent by mass. [it] Detemine the molality of chloroform in the sater sample.

Part (i): Calculating the percentage by mass ppm by mass is the amount of solute in 10 6g of solution. Given: Concentration =15ppm Now, 106g of solution contains =15g of CHCl 3 Hence; 100g of solution contain =15/106×100 =15×10 −4=0.0015% Part (ii): Calculating the molality of chloroform Molar mass of chloroform =119.5g/mol Mass of solvent =1000000−15=999985g Molality=Molesofsolute/Mass of solvent×1000 m=15119.5×999985×1000 m=1.266×10 −4m

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298K. Calculate: (1) molar mass of the solute (2) vapour pressure of water at 298K. 100 g of liquid A (molar mass 140 g mol −1) was dissolved in 1000 g of liquid B (molar mass 180 g mol −1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr. Vapour pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot p total’ p chloroform’ and p acetoneas a function of x acetone. The experimental data observed for different compositions of mixture is. 100× x acetone 0 11.8 23.4 36.0 50.8 58.2 64.5 72.1 p acetone /mm Hg 0 54.9 110.1 202.4 322.7 405.9 454.1 521.1 p chloroform /mm Hg 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7 The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298Kif the Henry’s law constants for oxygen and nitrogen are 3.30 × 10 7mm and 6.51 × 10 7mm respectively, calculate the composition of these gases in water. Shaa...

(i) 1 ppm is equivalent to 1 part out of 1 million (10 6) parts. ∴ Mass percent of 15 ppm chloroform in water `=15/10^6xx100` `≃ 1.5 xx10^-3 %` (ii) 100 g of the sample contains 1.5 × 10 –3g of CHCl 3. ⇒1000 g of the sample contains 1.5 × 10 –2g of CHCl 3. ∴ Molality of chloroform in water `=(1.5xx10^-2 "g")/("Molar mass of" "CHCl"_3)` Molar mass of CHCl 3= 12.00 + 1.00 + 3(35.5) = 119.5 g mol –1 ∴ Molality of chloroform in water = 0.0125 × 10 –2m = 1.25 × 10 –4m