Question

5. What is the mass of solute in each of the following solutions?

A.) 2.95 L of 0.200 M FeCl3

B.) 2.95 L of 0.200 M KIO4

C.)60.0 mL of 0.335 M ZnSO4

D.)60.0 mL of 0.335 M Ni(NO3)2

4. What volume of each of the following solutions contains the indicated amount of dissolved solute?

C. ) 50.0 g solute in 0.441 M K2CO3

D.) 50.0 g solute in 0.441 M Fe(ClO3) 3

Answer 1

5)

A)

volume , V = 2.95 L

use:

number of mol,

n = Molarity * Volume

= 0.2*2.95

= 0.59 mol

Molar mass of FeCl3,

MM = 1*MM(Fe) + 3*MM(Cl)

= 1*55.85 + 3*35.45

= 162.2 g/mol

use:

mass of FeCl3,

m = number of mol * molar mass

= 0.59 mol * 1.622*10^2 g/mol

= 95.7 g

Answer: 95.7 g

B)

volume , V = 2.95 L

use:

number of mol,

n = Molarity * Volume

= 0.2*2.95

= 0.59 mol

Molar mass of KIO4,

MM = 1*MM(K) + 1*MM(I) + 4*MM(O)

= 1*39.1 + 1*126.9 + 4*16.0

= 230 g/mol

use:

mass of KIO4,

m = number of mol * molar mass

= 0.59 mol * 2.3*10^2 g/mol

= 1.357*10^2 g

Answer: 136 g

C)

volume , V = 60 mL

= 6*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.335*6*10^-2

= 2.01*10^-2 mol

Molar mass of ZnSO4,

MM = 1*MM(Zn) + 1*MM(S) + 4*MM(O)

= 1*65.38 + 1*32.07 + 4*16.0

= 161.45 g/mol

use:

mass of ZnSO4,

m = number of mol * molar mass

= 2.01*10^-2 mol * 1.614*10^2 g/mol

= 3.245 g

Answer: 3.24 g

D)

volume , V = 60 mL

= 6*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.335*6*10^-2

= 2.01*10^-2 mol

Molar mass of Ni(NO3)2,

MM = 1*MM(Ni) + 2*MM(N) + 6*MM(O)

= 1*58.69 + 2*14.01 + 6*16.0

= 182.71 g/mol

use:

mass of Ni(NO3)2,

m = number of mol * molar mass

= 2.01*10^-2 mol * 1.827*10^2 g/mol

= 3.672 g

Answer: 3.67 g

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