Molality : It is defined as the number of moles of the solute present in 1 kg of the solvent. It is denoted by 'm'. Molality(m)="Number of moles of solute"/"Number of kilo-grams of the solvent" Let w_A grams of the solute of molecular mass m_A be present in w_a grams of the solvent, then Molality (m)=w_A/(m_Axxw_B)xx100 Realtion 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_Axxm_B)/(w_Bxxm_A) (X_Axx1000)/(X_Bxxm_B)=(w_Axx1000)/(w_Bxxm_A)=m or (X_Axx1000)/((1-X_A)m_B)=m What is the molality of final solution obtained in the above problem

Molality : It is defined as the number of moles of the solute present

1.	Molality : It is defined as the number of moles of the solute present in 1 kg of the solvent. It is denoted by 'm'. Molality(m)="Number of moles of solute"/"Number of kilo-grams of the solvent" Let w_A grams of the solute of molecular mass m_A be present in w_a grams of the solvent, then Molality (m)=w_A/(m_Axxw_B)xx100 Realtion 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_Axxm_B)/(w_Bxxm_A) (X_Axx1000)/(X_Bxxm_B)=(w_Axx1000)/(w_Bxxm_A)=m or (X_Axx1000)/((1-X_A)m_B)=m What is the molality of final solution obtained in the above problem
Answer» 1.422 14.22 15.22 None SOLUTION :MOLE of SOLUTE=`(300xx0.3+500xx0.4)/40=7.25` MASS of solvent=800-290=510 MOLALITY`=7.25/510xx1000=14.22`