Question

Calculate 'm' of all the ions presents in the solution of $A{L}_2{\left(S{O}_4\right)}_3$ for $1M$ solution. Given $d_{solution}=2.342gm/mL$

• A. $2.5m$
• B. $1.25m$
• C. $1.3m$
• D. $1.2m$

Answer 1

The correct option is A $2.5m$

density of solution = 2.342 gm/ ml

so, mass of solution = 2342 g

in 1 M solution of $A{l}_2(S{O}_4{)}_3$ , ions present are 2 $A{l}^{3+}$ and 3$S{O}_4^{2-}$

so, total mass of solute = 56 + 288 = 340g

mass of solvent = 2342 - 340 = 2002 g

molality of all ions = $\frac{2}{2002}\ast 1000+\frac{3}{2002}\ast 1000$

so, molality = 2.49 m