The density of 1 M solution of NaCl is 1.0585 g mL^-1. The molality of the solution is

Molarity is the number of moles of solute present in one L of the solution.

$\mathrm{Molarity}=\frac{\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{Volume} \mathrm{of} \mathrm{solution} (\mathrm{in} \mathrm{L})}$

Molarity of NaCl is 1 M which depicts that one mole of NaCl present in one liter of the solution. Mass of NaCl present in one mole of NaCl is:

$\begin{array}{ccc}\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{NaCl}& =& \frac{\mathrm{Mass} \mathrm{of} \mathrm{NaCl}}{\mathrm{Molecular} \mathrm{weight} \mathrm{of} \mathrm{NaCl}}\\ 1 \mathrm{mol}& =& \frac{\mathrm{Mass} \mathrm{of} \mathrm{NaCl}}{58.5 \mathrm{g}/\mathrm{mol}}\\ \mathrm{Mass} \mathrm{of} \mathrm{NaCl}& =& 58.5 \mathrm{g}\end{array}$

Density is the ratio of mass and volume.

$\mathrm{Density}=\frac{\mathrm{Mass}}{\mathrm{Volume}}$

1.0585 g mL^-1 is the density of 1 M of NaCl solution.

$\begin{array}{ccc}\mathrm{Density}& =& 1.0585 \mathrm{g}/\mathrm{mL}\\ & =& 1.0585 \mathrm{g}/\mathrm{mL}\times \frac{1000 \mathrm{mL}}{1 \mathrm{L}}\\ & =& 1058.5 \mathrm{g}/\mathrm{L}\end{array}$

Mass of solution in 1 L is 1058.5 g.

Mass of solution is equal to the mass of water (solvent) and the mass of NaCl (solute).

$\begin{array}{ccc}\mathrm{Mass} \mathrm{of} \mathrm{solution}& =& \mathrm{Mass} \mathrm{of} \mathrm{NaCl}+\mathrm{Mass} \mathrm{of} \mathrm{water}\\ 1058.5 \mathrm{g}& =& 58.5 \mathrm{g}+\mathrm{Mass} \mathrm{of} \mathrm{water}\\ \mathrm{Mass} \mathrm{of} \mathrm{water}& =& 1000 \mathrm{g}\end{array}$

Molality is the number of moles of solute present in volume of solvent (in kg).

$\begin{array}{ccc}\mathrm{Molality}& =& \frac{\mathrm{Number} \mathrm{of} \mathrm{moles}}{\mathrm{Volume} \mathrm{of} \mathrm{solvent} (\mathrm{in} \mathrm{kg})}\\ & =& \frac{1 \mathrm{mol}}{1000 \mathrm{g}\times {\displaystyle \frac{1 \mathrm{kg}}{1000 \mathrm{g}}}}\\ & =& 1 \mathrm{m}\end{array}$