Question

The density of 1M solution of NaCL is 1.0585 g/L. The molality of the solution is
A. 1.0585
B. 1.00
C. 0.10
D. 0.0585

Answer 1

Hint: Molality is defined as the number of moles of solute present in 1 kilogram of solvent. Below is the formula for molality-
  \[m = \dfrac{{moles\;of\;{\text{NaCl}}}}{{{\text{weight of solvent}}}}\]
The SI unit for molality is mol/kg

Complete step by step answer:
A solution with a molality of 3 mol/kg is often described as 3 molal or 3m “However following the SI unit system of units, mol/kg or a related SI unit is now preferred.
In the cases of solutions with more than one solvent molality can be defined for the mixed solvent considered as a pure-pseudo-solvent. Instead of mole solute per kilogram solvent as in the binary case, units are defined as mole solute per kilogram mixed solvent.
Now according to question
\[m = \dfrac{{moles\;of\;{\text{NaCl}}}}{{{\text{weight of solvent}}}}\]
Weight of solvent = Weight of solution – Weight of NaCl.
= 1.0585 × 1000 – 58.5
= 1058.5 – 58.5 = 1000g = 1 kg.
Therefore, \[m = \dfrac{1}{1} = 1\]
Hence the molality of the solution is 1.

Note:
 Molarity is the number of moles of solute present in 1 litre solution and molality is the number of moles of solute present in 1 kilogram of solvent.Molality is preferred over molarity as molality is independent of temperature.