You want to prepare 50cm of 0.1m naoh from exactly 50% (wt/wt) naoh solution. what volume of the 50% solution is required

We want 50 cm^3 of 0.1 M NaOH, i.e we need (50 ml /1000 mL) * (0.1 mole)
= 0.005 moles of NaOH >>> (1)
we have 50 % (wt/wt)NaOH, i.e we have 50 g NaOH in 100 Solution
= 50 g / 100 g solution, to convert this to molarity (mol/L) we will do that:
(50g NaOH* 1.5253 density of solution(g/ml)* 1000)/( 40 M.wt NaOH(g/mol)*100g Solution) = 19.07 mol/L >>> (2)
From (1) and (2), ∴ 0.005 moles of NaOH = 19.07 mol/L
∴ the Volume (L) of 50% of solution = 0.005 moles / 19.07 mol/L
                                                            =  2.6 * 10 ^ -4 L = 0.26 mL

Iwould you figure it out but you don't have a picture or in your question as to what the physical property you're looking for is

The air is going to come out because of the low pressure.