Question

7. How many millilitres of 3.50 M NaOH can be prepared from 75.00 pared from 75.00 grams of solid NaOH? 8. A stock solution of 2.00 M CuCl2 is available. What volume of this stock solution is needed to prepare 250.0 mL of a 0.100 M CuCl2 solution? 9. A 4.691 g sample of MgCl2 is dissolved in enough water to give 750 mL of solution. What is the magnesium ion concentration in this solution? A) 3.70 x 10-2 M D) 4.93 x 102 M B) 1.05 x 10-2 M E) 0.131 M C) 6.57 x 10-2 M

Answer 1

Answer

ма. w Na on a 15 molar mass of Naon = 39.99 g mal no of moles of Nads & mass of woon molar mass of woon - 759 29.99 g mol = 1.875 mol molarity = 3.50M volume = ? No of moles of Naon = molarity & volume 1.875 mol a 3.50 moi x volume 1.875 mol - volume 3.50 molt 0.53514 - volume =) 535.7 me volume of 3.50 M Naoh is needed 11 L = 1000 me) & using molarity equation, we have a - MIUI (stock s01") = Mahal biluted solution) 2.00 M xv = 0./00 M X anoml = d.looM X 2.00M 50 me N = 12.5 me v = 12.5 me stock solution V dume of needed & 1215 me