What is the "molality" of a 30*g mass of sodium hydroxide dissolved in a 100*g mass of water?

$\text{Molality} = 7.5 \cdot m o l \cdot k {g}^{-} 1$
$\text{Molality"="Moles of solute"/"Kilograms of solute}$
And here we have $30 \cdot g$ $N a O H$ solute..........
So $\text{Molality} = \frac{\frac{30 \cdot g}{40.00 \cdot g \cdot m o {l}^{-} 1}}{100 \cdot g \times {10}^{-} 3 \cdot g \cdot k {g}^{-} 1} = 7.5 \cdot m o l \cdot k {g}^{-} 1.$