Question

You have to make 650 g of 0.68 molal aqueous potassium permanganate (KMnO4, M.Wt 158.0) solution. What mass (9) of KMnO4 do you need to weigh (two decimal places)? Answer:

Answer 1

Given: weight of solution (W) = 650 gm molality of solution ( = 0.68 mol. Wit of kmn Oy = 158.0 Suppose, moles of km nou required =n molality is given by molality = moles of KMnoy x 1000 Mass of water (gm)

m weknow, Suppose mass of water = m gm. So, putting the value in the above equation 0.68 = n X1000 (1) Mass of KMnOy = moles X_Molinet. nx.158_ . — 12) weight of (kmmoy t Water) = weight of solution ; 158nt m = 650 - 12 from equation (1) and (3) 1589 + looon = 6.50 0.68 158 n + 147059 n = 650 > 1628.59 n = 650 - n = 0.39g ~ oiy moles from equation (2)_ I mess of KMnOy - 158x9_3158X .4 = 63.gm ans