Question

Find the molarity of individual ions in the following solution : 44.0mL of 0.35M NaCl and 25.0 mL of 0.24M CaCl₂ are mixed together.

Answer 1

Given Nach 44ml, 0.35M Cacl5 25ml, 0.2419 we know that number of moles = concentration x Volume (int) n = Miu (inn) Nacht Naci moles, M=0.35 Molleit V = 44m1 = 44 x1631 .. h= (0.35 molait) x ( 44x16°) = 15.4 x 15 mol = 1.54x162 mol n = 1.54 x 16 mol cacl, : V=25ml = 25X10°L, M = 0.24 molleit ne (25x16% eit) x (0.24 moleit) I 6X163 mol Jour 91x9=u

Naci Nat +cí cach - cat? + aci In the solution total ions are 1 nat, 7 cath, a ci Nach Nat + ci 1-54800 a mol so wat = ci = 1.54X 10 mol. cach → 6x16mol catha taci [ ] Grid 6 X to moi RX 6X10 mol = 12 x 10 mol. concentration a number of moles Volume in L • Nat] . 1:544 x10 a mol 0.2231 69x163ī - Jwaty c0.223M Total Volleye =Rs+44 June V=69m =698163 [cat 2] = 6X163 mol. 69 x163 - 0.087 mole [cate] = 0.0877

| c ] moles! 1.54x 10 mol + 12x10-3 mol (1.54 + 1-2)x162 mol = 2.74x16 mol 2.74x10-7 mol 69x103 - :: M = - 0.397 M .(CT) = 0.397M