How many grams of pure naoh must be used to prepare 10.0 l of a solution that has a ph of 13?

The concentration of OH- ( symbol: [OH-] ), is equal to 10^-pOH (ten to the pOH'th power). The pOH equals 14 minus the pH, because the pH + the pOH = 14. So the pOH is 14-13= 1. Now the concentration of OH- is 10^-1 (= 1) moles/Litre 

NaOH (s) --> Na+ (aq) + OH- (aq) 
1. : 1. : 1 

So by dissolving one mole of NaOH, you get one mole of Na+ and one mole of OH-. Meaning that the molarity (number of solved NaOH in one Litre) of NaOH is 1 mole/Litre, because the ratio is 1:1. This means, in ten litres of water there are also ten moles of NaOH. And the weight of one mole of NaOH is 40.00 grams (look it up in literature). So in ten litres solution with a pH of 13, there are 40.00*10 = 400 = 4 * 10^2 grams of NaOH dissolved

The answer is d.potassium