The molality of 1M solution of NaCl (specific gravity $1.0585g/ml$) is :

$(a)\;1.0585\qquad(b)\;1.0\qquad(c)\;0.10\qquad(d)\;0.0585$

$m=\large\frac{Mole\;of\;Nacl}{Weight\;of\;solvent\;in\;kg}$
Weight of solvent =weight of solution-weight of Nacl
$\Rightarrow 1.0585\times 1000-58.5$
$\Rightarrow 1058.5-58.5=1000g=1kg$
$\therefore m=1/1$
$\Rightarrow 1$
Hence (b) is the correct answer.
edited Oct 28, 2013