Question

1. When 5.0 g CaCl2 is dissolved in enough water to make a 0.500 L solution, what is the molarity of ions in solution?

2. How many formula units are in 0.25 mole of Na2O?

3. How many nitrate ions are present in 0.200 mol of Zn(NO3)2?

Answer 1

1)

Molar mass of CaCl2,

MM = 1*MM(Ca) + 2*MM(Cl)

= 1*40.08 + 2*35.45

= 110.98 g/mol

mass(CaCl2)= 5.0 g

use:

number of mol of CaCl2,

n = mass of CaCl2/molar mass of CaCl2

=(5 g)/(1.11*10^2 g/mol)

= 4.505*10^-2 mol

volume , V = 0.5 L

use:

Molarity,

M = number of mol / volume in L

= 4.505*10^-2/0.5

= 9.011*10^-2 M

This is concentration of CaCl2

1 molecule of CaCl2 gives 3 ions which are 1 Ca2+ and 2 Cl-

So,

[ions] = 3*[CaCl2]

= 3*9.011*10^-2 M

= 0.270 M

Answer: 0.270 M

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