An aqueous fo diabasic acid (molecular mass = 118) containing `35.4 g` of acid per litre of the solution has density `1.0077 g mL^(-1)`.
Express the concentration in as many ways as you can?

Answer

A, B, C, D

Explanation

`M = (35.4 xx 1000)/(118 xx 1000) = 0.3`
`m = (W_(2) xx 1000)/(Mw_(2) xx (V_(sol) xx d_(sol) - W_(2)))`
`= (35.4 xx 1000)/(118 (1000 xx 1.0077 - 35.4))`
`= (35.4 xx 1000)/(118 xx 972.3) = 0.31`
`N = n xx M = 2 xx 0.3 = 0.6`
Weight of solvent `(W_(1))`
= Weight of solution - Weight of solute
`= V_(sol) xx d_(sol) - W_(2)`
`= 1000 xx 1.0077 = 35.4 = 972.3`
`X_(2) = (n_(2))/(n_(1) + n_(2)) = (W_(2) // Mw_(2))/((W_(1))/(Mw_(1)) + (W_(2))/(Mw_(2)))`
`= (35.4 // 118)/((972.3)/(18) + (35.4)/(118))`
`= (0.3)/(54 + 0.3) = 0.055`
`X_(1) = 1 - X_(2) = 1 - 0.0055 = 0.9945`