(d)105 mL of 0.545 M C&Ci2; sonu 4. -/1 pointsZumintro6 15.P.047 -orKN03 is requ

Question

(d)105 mL of 0.545 M C&Ci2; sonu 4. -/1 pointsZumintro6 15.P.047 -orKN03 is required to prepare 200. mL of 0.291 M KNO3 solution? S. -/4 pointsZumintro6 15.P.049 the number of moles of the indicated ion present in each of the following solutions. (a) Na* ion in 1.00 L of 0.348 M Na2504 solution mol (b) Cl ion in 4.30 L of 0.11 M FeCl3 solution (c) NO3 ion in 180. mL of 0.58 M Ba(NO3)2 solution mol (d) NH4+ ion in 340. mL of 0.245 M (NH4)2504 solution mol net/web/Student/Assignment-Responses/last?dep-16951685 tt

Explanation / Answer

q1) molarity of solution = 0.291 M

molar mass of KNO3 = 101 g/mol

volume of KNo3 solution = 200mL

We know molarity = moles/L

= (mass/ molar mass) /V L

0.291 = (mass/ 101) /0.2

thus mass of KNow to be taken = 5.8782g

Q2)

a)moles of Na+ in 1L of 0.348M Na2 SO4

one mole of Na2SO4 has 2 Na+ ions.

0.348M means 1 L has 0.348 x 2 Na+ ions

thus moles of Na+ in 1L solution -=0.696 mol

b) Cl- in 4.30L of 0.11M FeCl3

FeCl3 gies 3 Cl- ions

Thus 1 L of solution has 3 x 0.11 moles of Cl-

4.3L of solution has = 4.3 L x 3 x 0.11mol/L

= 1.419 mol of Cl- ions.

c) moles of NO3- in 180mL of 0.58M Ba(NO3)2

thus

one mole of barium nitrate gives 2 NO3- ions

thus 1L of solution has 0.58x 2 moles of NO3- ions.

Thus 180mL = 0.18 L of solution has

= 0.18x 0.58 x 2 mol

= 0.2088 mol of NO3-

d) Nh4+ in 0.245 M (NH4)2SO4

One mole of ammonium sulfate has 2 moles of NH4+ ions

Thus 1L of solution has =2x 0.245 moles of NH4+ ions

Hnece 340mL =0.34 L of solution has

= 0.34 x 2 x 0.245

= 0.1666 mol of NH4+

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