Normality (N) (नॉर्मलता)|Molarity (M) (मोलरता)|Example|Formality (F) (फॉर्मलता)|Molality (m) मोललता)|Parts Per Million (ppm)|Mole Fraction (मोल प्रभाज)|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

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

Out of moalrity (M), molality (m), formality (F) and mole fraction (x) those independent of temperature are:

Concentration and concentration terms|Molarity|Normality|Formality|Parts per million

Out of molarity (M), molality (m), normality (N) and mole fraction (x), those independent of temperature are

The concentrations of soluitons can be expressed in number of ways , viz : mass fraction of solute (or mass percent), Molar concentration (Molarity ) and Molal concentration (molality). These terms are known as concentration terms and also they are related with each otehr i.e., knowing one concentration terms for the solution, we can find other concentration terms also. the definition of different cencentration terms are given below: Molarity : It is number of moles of solute present in one litre of the solution. Molality : It is the number of moles of solute present in one kg of the solvent. Mole fraction =("Mole of solute")/("Moles of solute" + "Moles of solvent") If molality of the solution is given as a, then mole fraction of the solute can be calculated by Mole Fraction =(a)/(a+(100)/(M_("solvent"))),=(axxM_("solvent"))/((a xx M_("solvent")+1000)0 where a=molality and M_("solvent") =Molar mass of solvent We can change : Mole fraction hArr Molality hArr Molarity What is the mole fraction of the solute?

A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M H_(2)SO_(4) . The residual acid required 60 mL of 0.5 M solution of NaOH for neutralization. Find the percentage composition of nitrogen in the compound. Strategy: Step 1 . Convert molarity into normality using the relation Normality (N) = n xx Molarity (M) where n factor is either the acidity of base or basicity of acid. Step 2 . Calculate the milliequivalents of NaOH which is equal to the milliequivalents of unreacted H_(2)SO_(4) . Step 3 . Calculate the milliequivalents of total H_(2)SO_(4) and subtract the milliequivalents of unreacted H_(2)SO_(4) to get the milliequivalents of NH_(3) evolved. Step 4 . Calculate the equivalents of NH_(3) , moles of NH_(3) , and moles of N . Step 5 . Calculate the mass of N in the organic compound. Step 6 . Finally, calculate % of N or directly apply Eq. (13.8) or (13.9) to get the % of N in the organic compound.

Molality : It is defined as the moles of the solute pressent in 1 kg of the solvent . It is denoted by m. Molality(m) =("Number of moles of solute")/("Number of kilograms of the solvent") let w_(A) grams of the solute of molecular mass m_(A) be present in w_(B) grams of the solvent, then: Molality(m) = (w_(A))/(m_(A)xxw_(B))xx1000 Relation between mole fraction and molality: X_(A)=(n)/(N+n) "and" X_(B)=(N)/(N+n) (X_(A))/(X_(B))=(n)/(N)=("Moles of solute")/("Moles of solvent")=(w_(A)xxm_(B))/(w_(B)xxm_(A)) (X_(A)xx1000)/(X_(B)xxm_(B))=(w_(A)xx1000)/(w_(B)xxm_(A))= m or (X_(A)xx1000)/((1-X_(A))m_(B))=m What is the quantity of water that should be added to 16 g methonal to make the mole fraction of methonal as 0.25?

Choose the correct answer among the following is (i) Normality and molarity of HCl are same (ii) Molality and molarity are independent of temperature (iii) Normality of 1M KMnO_(4) in acidic medium is 5N.