Question

A solution
has
is 27% (m/m) of, H2SO4
has a density of 1.64gml.Calculate
the molarity and mole friction
of the solution.​

Answer 1

Answer:

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Answer 2

Solution:-

• 27 % m/m  means 27 g of solute(H2SO4) in 100 g solution

Solvent(H20) = Solution - solute

• Solvent(H20) = 100 - 27 = 73 g

H2SO4

• mass of solute(w) = 27 g

• Molar mass (M) = 2(H) + S+4(o) = 2 + 32 +64= 98  g

we know that ,

⟹ n = w / M

⟹ n = 27 /98

⟹ n = 0.275

H2O

• Mass of solute(w) = 73 g

• Molar mass (M) = 2(H) +O= 2 +16= 18  g

we know that ,

⟹ n' = W '/ M '

⟹ n '= 73 /18

⟹ n ' = 4.05

Mole fraction of solute

⟹ X = n / n + n '

⟹ X = 0.275 / 0.275 + 4.05

⟹ X = 0.275 / 4.325

⟹ X = 0.07

⟹ X = 0.1

Mole fraction of solvent

⟹ X ' = n' / n + n '

⟹ X ' = 4.05 / 0.275 + 4.05

⟹ X'  = 4.05 / 4.325

⟹ X'  = 0.9

Mole fraction of solution = 0.1 +0.9 = 1

Remember : The mole fraction of the solution will always come out to be 1

Molarity

As per the formula ,

⟹ M = n / V (L)

here ,

• M = molarity
• n = moles of solute
• V = volume of solution

Density of the solution(D) = 1.64 g / ml

Mass of the solution (V)= 100g

we know that ,

⟹ D = M /V

⟹ V = M /D

⟹ V = 100 / 1.64

⟹ V = 60.97 ml

⟹ V = 0.06 L

Now let's substitute the values of V and n in the above equation

⟹ M = 0.275 / 0.06

⟹ M = 4.58 M

The molarity of the solution = 4.58 M