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!