Question

4.56 How many atoms (or ions) of each element are in 140.0 g of the following substances? (a) BaSO4 (c) 02 (b) Mg3(PO4)2 (d) KBr

Answer 1

Note, for such questions you need to find molecular weight of the compounds or it may be given in texts.
Solution. The following question is based on Mole concept. One thing you should known is that, 1 mole of any substance compound contains 6.022 X1623 atoms molecules. also, Imd = Molecular weight of substance compound. for example, 32g of Q = 1 mole of Oq = 6:022 x 1023 نه of بقلبهاهم | Now, to find out atoms/or ions in case of ionic compound) - first we need to see or to find out number of moles of that compound, and then we will use 1 mole = 6.022 x 1023 molecules atomslions, One thing which also you take care in case of ionic - compounds (BaSO4, Mg3(PO4), KBr) is the number of ions on dissociation of single molecule, we need to multiply accordingly, @ Basof, weight=1400g Molecular weight (Mor Wt.) = 233-38g/mol of BaSO4 i humber of moles of Bas Oq un 14ogous- na 140.09 233.383/mol = 0.59 mol hus Total Number of Baso molecules un 0.59 mol = 0.69x6.022 x 1623 molecules = 3.6 % 1623 molecules of Basoa

Now, BaSO4 will dissociate as Csince it is lohics Basoa = Ba2+ +504 Thus, humber of Ba2tions = number of soz ions = 3.6 X 1023 O MGB(204), Mot.ct=262.859Imol. n = 140.09 50.53 moles. 262.83gimo) -0.53% 6.022x123 mobecules = 3.2% 1023, Mg 3(PO4)2 molecules 1. also, Mg2 (0)2 - 3Mg2+ + 2 988 Thus, number of Mq2tions= 343.2*2223 59.6x1023 similarly, number of Posions = 2X 3.2 x 1023 = 6.4 x 1023 n=140.0g = 4.375 molest of 2 = 32g/mol | 32.09 i humber of molecules of O₂ = 4.375X6.022 X 1623 = 2.6 X 1024, 0, melecules. Os will not dissociate, as it is a coulent compound

Tâm L01, 140.0% , Mel-uo+. - 119.5 ing) n=140.0g = 1.176 moles of kBr 119.00291mel = 1.17646.822x 1023 molecules of KB? = 7.08 x 1023 molecules of KBs. also, KBr 2 kt + Br This on dissociation KBr, is giving out one molecule of Kt and Br each. So, number of ktiony = 7:08* 1023 and, number of Brions = 706x 1023.