[tex]M_{NaOH} = 23 + 16 + 1 = 40[/tex]
1 mol NaOH ............................ 40 g
x moli NaOH .......................... 300 g
[tex]x = \frac{300\cdot 1}{40} = \frac{30}{4} = 7.5[/tex]
[tex]c_M = \frac{moli}{volum} = \frac{7.5 moli}{60 l} = 0.125 M[/tex]