Answer:
Mass 1=3M
Mass 2=?
Volume1=75mL
Volume2=250mL
By using molarity formula:
<u>mass1*volume 1=mass2*volume 2</u>
3M*75=mass2*250
mass2=225/250
mass2:0.9M
<h3>the molarity of a solution is 0.9M.</h3>
Answer : The mass of calcium chloride (in g) needed is, 1.92 grams.
Explanation : Given,
Boiling point of elevation constant 
for water = 
Mass of water (solvent) = 
Molar mass of 
= 110.98 g/mole
Formula used :
where,
= change in boiling point = 
i = Van't Hoff factor = 3 (for electrolyte)
= boiling point constant for water
m = molality
Now put all the given values in this formula, we get
Therefore, the mass of calcium chloride (in g) needed is, 1.92 grams.
Answer:
Answer: Chemical change; Chemical property of heat of combustion
Explanation:
A physical change is defined as a change in which there is alteration in shape, size etc. No new substance gets formed in these reactions.
Explanation:
Answer:NH
+
4
:
0.272
M
SO
2
−
4
:
0.136
M
Explanation:
We can solve this problem using some molarity calculations:
molarity
=
mol solute
L soln
We should convert the given mass of
(NH
4
)
2
SO
4
to moles using its molar mass (calculated to be
132.14
g/mol
):
10.8
g (NH
4
)
2
SO
4
⎛
⎝
1
l
mol (NH
4
)
2
SO
4
132.14
g (NH
4
)
2
SO
4
⎞
⎠
=
0.0817
mol (NH
4
)
2
SO
4
This is the quantity present in
100
mL soln
, so let's calculate the molarity of the solution (converting volume to liters):
molarity
=
0.0817
l
mol (NH
4
)
2
SO
4
0.100
l
L soln
=
0.817
M
10
mL
of this solution is added to
50
mL H
2
O
, which makes a
60
-
mL
total solution.
We can now use the dilution equation
M
1
V
1
=
M
2
V
2
to find the molality of the new,
60
-
mL
solution:
(
0.0817
M
)
(
10
l
mL
)
=
(
M
2
)
(
60
l
mL
)
M
2
=
(
0.817
M
)
(
10
mL
)
60
mL
=
0.136
M
This means that there are
0.136
moles of
(NH
4
)
2
SO
4
per liter of solution.
Let's recognize that
1
mol (NH
4
)
2
SO
4
contains
2
mol NH
+
4
1
mol SO
2
−
4
The concentrations of each ion is thus
(
2
)
(
0.136
M
)
=
0.272
M
NH
+
4
(
1
)
(
0.136
M
)
=
0.136
M
SO
2
−
4
Explanation:
It is a solution where a solute cannot dissolve anymore into the solvent.
