Answer:
a. Ca2+ 1M ; NO3- 2M
b. Na+ 4M ; SO4-2 2M
c. NH4+ , Cl- 0,186M
d. K+ 0,056M ; PO4-3 0,0188 M
Explanation:
Here are the ionization equations
Ca(NO3)2 ----> Ca2+ + 2NO3-
Na2SO4 ------> 2Na+ + SO4-2
NH4Cl ------> NH4+ + Cl-
K3PO4 -----> 3K+ + PO4-3
Concentrations must be in Molarity (moles/L)
Ca(NO3)2 0,1 mole in 100mL
100mL = 0,1 L
Molarity Ca(NO3)2 1M (0,1 /0,1)
Ca(NO3)2 ----> Ca2+ + 2NO3-
1 M 1M 2M
Na2SO4 2,5 moles in 1,25L
Molarity Na2SO4 2,5 moles /1,25L = 2M
Na2SO4 ------> 2Na+ + SO4-2
2M 4M 2M
5.00 g of NH4Cl in 500mL
We have to resort to the molar mass
Let's make the rule of three
500mL ______ 5g
1000mL _____ (1000mL . 5g) / 500mL = 10 g
(These are the grams in 1L, so with the molar mass, we get the moles)
Mass / Molar mass = moles --> 10g / 53,5 g/m = 0.186 moles
NH4Cl ------> NH4+ + Cl-
0,186 M 0,186M + 0,186M
1.00 g K3PO4 in 250.0 mL
(The same as before)
250 mL ________1 g
1000 mL _______ (1000 mL . 1 g) /250mL = 4 g
These are the grams in 1L, so with the molar mass, we get the moles
Mass / Molar mass = moles --> 4g /212,26 g/m = 0,0188 moles
K3PO4 -----> 3K+ + PO4-3
0,0188 M 0,056M + 0,0188 M
