Question

3. -19.09 points ZumChemP8 4.E.026. Calculate the concentration of all lons present in each of the following solutions of strong electrolytes. 0.140 mole of Ca(NO3)2 in 100.0 mL of solution (a) Ca2+ NO3 5.0 moles of NazS04 in 1.25 L of solution (b) Na+ so,2- 5.00 g of NH4CI in 540.0 mL of solution (c) м NH4+ CI- 1.00 g K3PO4 in 200.0 mL of solution (d) к+ PO43- Need Help? Read It Supporting Materials

Answer 1

As given solution are strong electrolytes, so dissociate completely into ions. Thus we can find no of moles of ions from the moles of solution given.

And concentration of ions( Molarity )

= No. Of moles of that ion / volume of solution in L

Ca(NO3) - Cat + 2007 Holes 0.140 moles 0.140 0.140x2 moles volume of sol = 100ml concentration of Ca2+ - 0.140 Moles - 100x10-3 = 1.40 Mol/L concentration of NO3 = 0.140 x 2 100x10-3 = 2.80 MOLL Ca 2+ = 1.40M NO = 2.80M Na, sou 2Na+ + sou (6) Moles 5.0 Moles 5.082 Moles 5.0 Moles Volume of sol = 1.25L Conc. of wat = 5.0*2 moles. 1.25L es = 8 moth Conc of 80 - - 5.0 Holes 4.0MOL/L 1.25L Na + = 8M so = 40

(6) weight of why I given = 5.0g (1 Motor Mass of Nhy C - 53.59/rol No. of Moles of Nhy d = w 5.09 = 0.0935 Moles 53. 5 9 1 mol Nhy e Nhyt + le- 0.0935 0.0935 Moles 0.0935 Volume of soln - 540.00l conc. of NHY + = 0.0935 Holes es - 0.17M 540 X10 3 = UTA 0.17M conc. of ce- = 0.0935 moles - 540 x10-3L NHY + = 0.174, = 0.17M

(d) No. of roles of Kg roy - gino 219.97 9 IMA Moles 19. Kapoy roles 1/212.27 > K+ + poy? - 3/12.27 1/212.27 27 .27 Volume of solna 200 me Conc. of K+ = 3 moles 212.27 x200x/0-3, = 0.071 M Cone. of Poy = 0.023M 219.27X900X10-3 Ikt = 0.07M py 3- = 0.02M