How many moles of MgCl2 will be needed to add to 810 gram of water to prepare a solution having 0.1
mole fraction of MgCl2.
Answers
Answer:
860 g MgCl
2
Explanation:
For starters, we know that a
1-M
solution contains
1
mole of solute for every
1.0 L
of solution.
This implies that a
4.5-M
magnesium chloride solution will contain
4.5
moles of magnesium chloride, the solute, for every
1.0 L
of solution.
So all you have to do now is figure out how many moles of magnesium chloride would be equivalent in
2.0 L
of solution to
4.5
moles in
'
1.0
L
of solution.
?
.
moles MgCl
2
2.0 L solution
=
= 4.5 M solution
4.5 moles MgCl
2
1.0 L solution
Use cross multiplication to get
?
=
2.0
L solution
1.0
L solution
⋅
4.5 moles MgCl
2
This will be equal to
?
=
9.0 moles MgCl
2
So, you know that you can get a
4.5-M
magnesium chloride solution by dissolving
9.0
moles of magnesium chloride in enough water to have a total volume of
2.0 L
of solution.
To convert this to grams, use the compound's molar mass
9.0
moles MgCl
2
⋅
95.211 g
1
mole MgCl
2
=
860 g
−−−−−
The answer is rounded to two sig figs, the number of sig figs you have for your values.
Answer:
169.1 gram of MgCl2
Explanation:
molecular mass of mgcl2 =95
molecular mass of water =18
no. of moles in 1000 g =1000/18= 16 (approx.)
mole fraction = no. of moles of solute /no. of moles of solution
let no. of moles of mgcl2 be n
0.1=n/(n+16)
1.6+0.1n=n
1.6=0.9n
n=1.6/0.9
n=1.78 moles
1.78 moles = 1.78 moles * 95 g/mol(molar mass )=1.69.1 gram
answer =169.1 gram of MgCl2 have to be added in 1000 gram of water to form 0.1 mole fraction of mgcl2