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

What are the statements?

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.