a certain public water supply contains 0.10 ppb of chloroform .how many molecules of chloroform would be obtained in 0.478 ml drop of this water ?
Hi Garima,
It is given that there is 0.10 ppb of chloroform in a certain water supply, which means 10^9 g of water contains 0.10 g of CHCL3 ( because 1 billion = 10^-9)
Now volume of water = 0.478 mL (given)
Assume density of water to be 1 g/mL
Mass of the water can be calculated as, mass = density * volume
Therefore, mass = 0.478 ml * 1 g/mL = 0.478 g
Now, 10^9 g of water contains 0.10 g of CHCL3
Therefore, 1 g of water will contain 0.10 g / 10^9 g of CHCL3 = 10^-10 g of CHCL3
Therefore, 0.478 g of water will contain 0.478 * (10^-10) g of CHCL3
Now, no. of moles can be calculated using the formula = Given Mass/ Molecular Mass
Therefore no. of moles of CHCL3 = (0.478 * 10^-10) / 119.5 = 0.004 * (10^-10) = 4 * (10^-13)
(molecular mass of CHCL3 = 119.5 g)
Now one mole of any compound contains = Na no. of molecules
therefore 4 * (10^-13) moles of CHCL3 will contain 4* 10^-13 * Na no. of molecules
Therefore, 0.478 mL of this water will contain 4 * 10^-13 * Na molecules of CHCL3
Hope this helps!
