Question

( rightarrow ) given, ( bmod operatorname{lif}_{y} m=x ), Ma (Benzene) ( = )
mole ( int x operatorname{action} 1 x_{B}=0.4 )
To find value of ( x ).
We lanow, molakty, ( x=frac{x_{B} x 1000}{left(1-x_{R}right) times M_{A}} )
( x Rightarrow frac{400}{0.6 times 78}+frac{400}{46.8} geqslant 0.5 )

# molality (m) dl 10 UI and mole fraction in the laboratory at 24°C, you W11 find (1) Mole fraction (x) and molality (m) (2) Mole fraction (2x) and molality (2m) (3) Mole fraction (x/2) and molality (m/2) (4) Mole fraction (x) and molality (2m) If x molal solution of a compound in benzene has mole fraction of solute equal to 0.4. Then the value of x is (1) 4.2 (2) 8.5 (3) 3.2 (4) 5.1 Equal masses of CH,COOH and oxalic acid are dissolved separately in equal volumes of solutions, then the molarity (1) Will be equal for the two solutions (2) For CH2COOH solution molarity will be greater than that of oxalic acid solution 5. Aakash Educational Services Limited - Regd. Office: Aakash Tow

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