What will be the molarity of the solution of 2.5 grams of ethanoic acid (CH_(3)COOH) in 75 grams of benzene? 12 | Intermediate...

Calculate molality of 2.5g of ethanoic acid (CH_(3)COOH) in 75 g of benzene.

Benzene melts at 5.50^(@)C and has a freezing point depression constant of 5.10^(@)C dot" m^(-1) . Calculate the freezing point of a solution that contains 0.0500 mole of acetic acid , CH_(3)COOH , in 125 g of benzene if acetic acid forms a dimer in this solvent :

What is the percentage dissociation of 0.1 M Solution of acetic acid? [k_a(CH_3COOH) = 10^( -5)]

Pure benzene freezes t 5.3^(@)C . A solution of 0.223 g of phenylacetic acid (C_(6)H_(5)CH_(2)COOH) in 4.4 g of benzene (K_(f) = 5.12 Kkg mol^(-1)) freezes at 4.47^(@)C . From the observation one can conclude that :

What would be the molality of 60% (mass/mass) aqueous solution of CH_(3)COOH ? (Molar mass of CH_(3)COOH = 60 g mol^(-1) )

Molarlity: An acqueous solution contains 128g of mehanol (CH_(2) OH) in 108g of water. Calculate the molarity of the solution. Strategy: Convert the grams of CH_(3) OH to moles of CH_(3) OH express the mass of H_(2) O in kilogram and apply the definition of molality.