Question
What will be the molality of the solution containing $$18.25\,g$$ of $$HCl$$ gas in $$500\,g$$ of water ?
A.
0.1$$\,m$$
B.
1$$\,M$$
C.
0.5$$\,m$$
D.
1$$\,m$$
Answer :
1$$\,m$$
Solution :
$$\eqalign{
& {\text{Molality}}\left( m \right) \cr
& = \frac{{{\text{No}}{\text{. of moles of solute}}}}{{{\text{Mass}}\,\,{\text{of}}\,\,{\text{solvent}}\,\,\left( {{\text{in}}\,\,kg} \right)}} \cr
& {\text{Molality}} \cr
& = \frac{{{W_B} \times 1000}}{{{M_B} \times {W_A}\left( {{\text{in}}\,\,g} \right)}} \cr
& = \frac{{18.25 \times 1000}}{{36.5 \times 500}} \cr} $$
$$\,\,\, = 1\,m$$ $$\left( {\because \,\,{\text{Molar mass of }}HCl = 36.5\,g\,mo{l^{ - 1}}} \right)$$