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The idea here is that you need to figure out how many moles of magnesium chloride,
MgCl
2
, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.
As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.
c
=
n
V
So, how many moles of magnesium chloride must be present in the target solution?
c
=
n
V
⇒
n
=
c
⋅
V
n
=
0.158 M
⋅
250.0
⋅
10
−
3
L
=
0.0395 moles MgCl
2
Now determine what volume of the target solution would contain this many moles of magnesium chloride
c
=
n
V
⇒
V
=
n
c
V
=
0.0395
moles
3.15
moles
L
=
0.01254 L
Rounded to three sig figs and expressed in mililiters, the volume will be
V
=
12.5 mL
So, to prepare your target solution, use a
12.5-mL
sample of the stock solution and add enough water to make the volume of the total solution equal to
250.0 mL
.
This is equivalent to diluting the
12.5-mL
sample of the stock solution by a dilution factor of
20
.
Answer:
acidic
Explanation:
because the acid is strong whereas the base is weak.
The molarity of HCl is calculated as below
find the moles of mg used
moles = mass /molar mass
= 2.00g/24 g/mol=0.083 moles
write the reacting equation
Mg(s) +2HCl (aq)= MgCl2(aq) + H2(g)
by use of mole ratio between Mg to HCl which is 1 :2 the moles of HCl is therefore
= 0.083 x2 =0.166 moles
molarity of HCl = moles of HCl/ volume in dm^3
that is 0.166/50 x1000= 3.3 M of HCl
Alkali metals have only one electron in their outermost energy levels , so (they can loose one electron to) they form a mono positive ion only . That's why alkali metal do not from dispositive ions .
Hope that helps:)
