100g of solution of urea in water has 10 g of urea (60g/mol) . find the molality of solution
Answers
Answer:
2.05 M
Explanation:
Molarity is defined as the number of moles of solute dissolved in per litre of solution.
Molarity, M=
V
n
n=number of moles of solute=mass/molar mass
V=volume of solution
mass of urea (NH
2
CONH
2
)=120g
Molar mass of urea= 60g/mol
Mass of water=1000g
Density of solution. ρ=1.15 g/mL
n=120/60=2mol
m
solution
=m
solute
+m
solvent
(∵ Law of conservation of mass)
ρ
solution
×V
solution
=120+1000 (using m=ρ.V)
1.15×V
solution
=1120
V
solution
=974mL=0.974L
thus M=
V
n
=
0.974
2
=2.05M
Molarity of the solution is 2.05M
The molality of the solution is 1.84 m
Explanation :
Step 1 : Find moles of solute.
The solute in this case is urea. Let us find moles of urea.
mol =\frac{grams}{Molar Mass}mol=
MolarMass
grams
Molar mass of urea is 60.1 g/mol
We have 10 grams of urea.
Let us plug in these values to find moles.
mol =\frac{10g}{60.1g/mol} = 0.166 molmol=
60.1g/mol
10g
=0.166mol
We have 0.166 mol urea.
Step 2 : Find mass of solvent
Water is solvent here .
The mass of solution is given as 100 g.
Mass of solution = mass of solute + mass of solvent
100 g = 10 g + mass of solvent
mass of solvent = 100 g - 10 g = 90 g = 0.09 kg
Step 3 : Use molality formula
The molality of a solution is calculated using following formula
m = \frac{(mol) Solute}{(kg) Solvent}m=
(kg)Solvent
(mol)Solute
molality =\frac{0.166 mol}{0.09 kg}molality=
0.09kg
0.166mol
molality = 1.84mmolality=1.84m
The molality of the given solution is 1.84 m