Question

+-/4 points 4. Determine the molarity of each of the following solutions. (a) 1.505 mol NaCl in 1.548 L of solution (b) 0.468 g of H2SO4 in 1.00 L of solution М (c) 20.85 g of Al(NO.), in 1503 mL of solution M (d) 2.53 kg of CuSO 5 H,0 in 315 L of solution (e) 0.005717 mol of Cl, in 11.10 mL of solution 0.000848 g of C.H,NO, in 2.38 mL of solution (f) M

Answer 1

violarity - Number of males of solute a volune of solution in litre ED) Molaiti = 1.505 mil = 0-972 milli = 0.972m. T 1.SUOL b) Mass 4 H2504 = 0.4684, Molar mass of H2SO4 = go glmal Alinubes of noles - Melar mas les = Nass en goona - 0.4688 = 0.00477 mil 98ginal 004775 ml = 0.004775 M meurety 1.00 L ( Mass GL AL(NO3)2 = 20.854 Mcla mass of Al (NO2)2 = 23 Anal Alubes quotes of 41 (ND) = = 10:0939 nal 979 neel 0.0979 mol = 0.0651 M 3ml 10503L d) Mass of Casoya 5410 = 2453 14 = 2.53x18g = 25301 Molar nass 4 (il s0y=5420 = 249.6@gmol Nember Pirelles = 25304 of =1013 mol 249,68 Jhal Mclarity = 10:13 hel = 0.03216Mhi 31SL

ce Nemibes of nules of cl = 0.60 5717 Wat volume of schitét = lalo mi z 11110L = Ocelle Altrubas restes holars Molasetyolt 0.005717mal – 0515M (4) Mats & GHEN Q2 = 0.9808484 molar mass of C3H7 NO₂ = 89.09 gnol les 4 GHZNO = Osodo 64@4 = 9,51&xzowel B D479.09 What volume of scheiteon = 2.38ml = 2:3843 0.002324 holaséty 9.5/@xio al = 0.004. málft = 0.0044 Malasif 0.00228 L Numbe T