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Molality is defined as the number of mol...

Molality is defined as the number of moles of the solute per kg of the solvent.

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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 molality of the above solution?

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?

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 Percentage (weight/vol) of NaCl persent in the solution is :

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 If the mole fraction of a solute is changed from (1)/(4) "to" (1)/(2) in the 800 g of solvent then the ratio tof molality will be:

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 The mole fraction of the solute in the 12 molal solution of CaCo_(3) is :

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 120 gm of solution containing 40% by mass of NaCl are mixed with 200 gm of a solution containing 15% by mass NaCl. Determine the mass percent of sodium chloride in the final solutions.

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 molarity of solutions if density of solution in 1.6 gm/ml?

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?

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 The molality of 1 litre solution with y% by (w/v) pf CaCO_(3) is 2 . The weight of the solvent present in the solution is 900g , then value of y is : [Atomic weight : Ca=40, C=12 , O=16]