Question

Find the molality and molarity of a 15% solution of H2So4 (density of H2So4_1.10gcm ,molar mass of h2so4_98)

Answer 1

15% H₂SO₄ solution means 15 g of H₂SO₄ is present in 100 g of Water.
No. of moles of H₂SO₄ = 15 g / 98 g/mol = 0.153 mol
Weight of solvent = 100g - 15g = 85g = 85 × 10^-3 kg

Molality = Moles of solute / Weight of Solvent in kg
= 0.153 / (85 × 10^-3)
= 1.8 m
Molality of given solution is 1.8 m ……[1]

Volume of solution = Mass of solution / Density
= 100 g / 1.1 g/cm^3
= 90.90 cm^3
= 0.090 Litres

Molarity = Moles of solute / Volume of solution in litres
= 0.153 / 0.090
≈ 1.7 M
Molarity of given solution is 1.7 M ……[2]

Answer 2

$\fbox{\texttt{\green{Correct\:Question:}}}$

Find the molality and morality of a 15% solution of H2SO4 (density of H2SO4 = 1.020 g cm^-3). (Atomic mass : H = 1, O = 16, S = 32 a.m.u).

$\fbox{\texttt{\pink{Given\::}}}$

15% solution of $\bold{H_2SO_4}$ means 15 g of $\bold{H_2SO_4}$ are present in 100 g of the solution, i.e.,

Mass of $\bold{H_2SO_4}$ dissolved = 15 g

Mass of the solution = 100 g

Density of the solution = $\bold{1.02\:g/cm^3}$

$\fbox{\texttt{\orange{To\:find:}}}$

a. Molality

b. Morality

$\fbox{\texttt{\red{Solution\::}}}$

★ Calculation of molality :

Mass of solution = 100 g

Mass of $\bold{H_2SO_4}$ (solute) = 15g

Mass of water (solvent) = $\\\bold{100-15=85\:g=\frac{85}{1000}\times{kg}=0.085\:kg}$

Molar mass of $\bold{H_2SO_4=98\:g\:mol^{-1}}$

$\bold{\therefore 15\:g\:H_2SO_4=\frac{15g}{98\:mol^{-1}}=0.153\:moles}$

$\bold{Molality=\frac{No.\:of\:moles\:of\:solute}{Mass\:of\:solvent\:in\:kg}}$

$\implies\bold{\frac{0.153\:mol}{0.085\:kg}=1.8\:mol\:kg^{-1}=1.8\:m}$

★ Calculation of morality :

$\bold{15\:g\:of\:H_2SO_4=0.153\:moles(calculated\:above)}$

$\bold{Volume\:of\:solution=\frac{Mass\:of\:solution}{Density\:of\:solution}}$

$\implies\bold{\frac{100}{1.02}=98.04\:cm^3=\frac{98.04}{1000}\times{L}=0.09804\:L}$

$\bold{Molarity=\frac{No.\:of\:moles\:of\:the\:solute}{Volume\:of\:solution\:in\:litres}}$

$\implies\bold{\frac{0.153\:mol}{0.09804\:L}=1.56\:mol\:L^{-1}=\:1.56\:M}$

Hence,

a. The molality of the solution is = 1.8 M

b. The morality of the solution is = 1.56 M