InterviewSolution
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InterviewSolution
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The relation between molarity `(C )` and molality `(m)` is given by )d= density of solution, `M` =molar mass of solute):A. `m=(1000 CM)/(1000 C-d)`B. `m=(1000 C)/(1000 d-CM)`C. `m=(Cd)/(1000d-CM)`D. `m=(CMd)/(1000-dM)` |
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Answer» Correct Answer - 2 `Molality (C )=n_("solute")/V_(L)` `"Molality "(m)=n_("solute")/(kg_("solvent"))` Thus, `m/C=V_(L)/(Kg_("solvent"))` `Kg_("solvent")= mass of solution in kg - mass of solute in Kg` `=[(("density of"),("solution in Kg "L^(-1)))(("volume of"),("soln in "L))]` `-[(n_("solute")xx" Molar mass of solute"),("in kg "mol^(-1))]` `=[d(kg L^(-1))V_(L)]-[C.V_(L)xxM/1000]` where M is the molar mass of solute in g and 1000 is the conversion factor to get molar mass in kg. Finally `Kg_("solvent")=V_(L)[d-(CM)/1000]` Substituting this result in Equlation (1), we get `m/C=V_(L)/(V_(L)[d-(CM)/1000])=1/([d-(CM)/1000])` `m=C/(d-(CM)/1000)` `=C/((1000 d-CM)/1000)` `m=(1000 C)/(1000 d-CM)` where m= molality in mol `Kg^(-1)` C= molarity in mol `L^(-1)` d= density of solution in `Kg L^(-1)` M= molar mass of solute in `g mol^(-1)` If molar mass of solute is expressed as `Kg mol^(-1)` then 1000 will be replaced by 1, that is, `m=C/(d-CM)` |
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