Question

c. A 15.0% stock solution of NaOH has a density of 1.13 g/mL. How would you prepare 250.0 mL of a 0.150 M solution of NaOH from this stock solution? (3 pts)

Answer 1

Naolt • g/m (② 100g ow since density 15.6% Stock solution of means Naot solution contain 1500 g Naolt of Naoh solution I Naot solution .: Volume 100 of g = 100g 1-13 g/m my - 88.5 mi Naolt Now, molar mass of of Naoh in =40g/mol ... Number mole 15.09 of Naot x 15.0 gx I mos 0.375 anos noog Naol . According to problem Accor to 88.5 my naolt solution contain 0.375 mol 0.375 88 Сл 1000 u u 0.375X1oua 88.5 - 6.24 .. Molarity of stock wash solution is 4,24 M

Now using vi xc, = V2 x 2 where, via gnitial volume C = Guitial concentration V2 Final valu olume Cra Final concentration here, v, -2, C,= 424M, V2 = 2500 me, e2= vix unggaa 2500 mx oisogh KV, 0.15OM asoomi xoo iso 4.24 8.84 mm of r. To fore fare asome are 2 sony OSOM Naot solution 8.84 mi stock Naolt solution should be added with (250 - 8184 ml = 24116 my distilled water