Parts Per Million(PPM) || Parts Per Billion (PPB) ||Solved Numerical Question || Conversion OF 1 Unit into Another || Meaning OF Molal/Semi-molal/Deci-molal/Centi-molal

Questions|Parts Per Million (PPM) And Parts Per Billion (PPB)|Summary

Mole Fraction|Questions|Parts Per Million (PPM) / Parts Per Billion (PPB)|Questions|Summary

Mole Fraction|Exercise Questions|Parts Per Million (PPM) / Parts Per Billion (PPB)|Exercise Questions|Summary

Mole Fraction |Exercise Questions|Parts Per Million (ppm) / Parts Per Billion (ppb)|Exercise Questions|Summary

Units of parts per million (ppm) or per billion (ppb) are often used to describe the concentrations of solutes in very dilute solutions. The units are defined as the number of grams of solution per million or per billion gram of solvent. Bay of Bengal has 2.1 ppm of lithium ions. What is the molality of Li^(+) in this water ? (Li = 7)

To find out the concentration of S O_2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below: Find the mean concentration of S O_2 in the air.

In a binary solution, the component present in smaller proportion is called solute while the one in excess is known as solvent. The solution containing 1 mole of the solute in 1L of solution is known as one molar solution while the solution in which 1 mole of solute is dissolved in 1 kg of the solvent is called one molal solution. The ratio of the no. of moles of a particular component to the total no. of moles in the solution is known as its mole fraciton The unit of molarity are

when 1.80 g of a nonvolatile solute is dissolved in 90 g of benzene, the boiling point is raised to 354.11K . If the boiling point of benzene is 353.23K and K_(b) for benzene is 2.53 KKg mol^(-1) , calculate the molecular mass of the solute. Strategy: From the boiling point of the solution, calculate the boiling point elevation, DeltaT_(b) , then solve the equation DeltaT_(b)=K_(b)m for the molality m . Molality equals moles of solute divided by kilograms of solvent (benzene). By substituting values for molality and kilograms C_(6)H_(6) , we can solve for moles of solute. The molar mass of solute equals mass of solute (1.80 g) divided by moles of solute. The molecular mass (in amu) has the same numerical value as molar mass in g mol^(-1) .

A 1 molal K_(4)Fe(CN)_(6) solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute A. The molar mass of A is _____u. (Round off to the Nearest Integer). [Density of water = 1.0 "g cm"^(-3) ]