Question

A solution is prepared by dissolving $20g\, NaOH$ in $1250 \,mL$ of a solvent of density $0.8\, g/mL$. Then the molality of the solution is

• A. $0.2\, mol \,kg^{-1}$
• B. $0.08\, mol \,kg^{-1}$
• C. $0.25\, mol \,kg^{-1}$
• D. $0.0064\, mol \,kg^{-1}$
• E. $0.5\, mol \,kg^{-1}$

Answer 1

Answer: (E)

see you have $20$ gram $NaOH$.
and molar mass of it is $=40$ gram.
so no. of moles $=20 / 40=1 / 2$ mole.
now volume of solvent $=1250\, ml$ and density $=0.8 \,g / ml$.
so mass of solvent $=0.8 \times 1250=1000$ gram $=1\, kg$.
so molality = mole of solute/ mass of solvent in $kg$.
$=1 / 2 / 1=1 / 2$ molal $=0.5$ molal.