Question

find the molarity if 10% of weight by weight aqua solution of H2SO4.If density of solution is 1.1 gram per ml

Answer 1

$\huge\underline\green{\sf Answer :-}$

$\large{\boxed{\sf Molarity (M)=1.12 Mol/L}}$

$\huge\underline\green{\sf Solution :- }$

$\large{\boxed{\sf Mass= Concentration \times Density \times Volume}}$

Given :-

Mass = ?

% Concentration = 10 % => 0.1

Density = 1.1 g/ml

Volume = 1000 ml

ON PUTTING VALUE :-

$\large\implies{\sf 0.1 \times 1.1 \times 1000}$

$\large{\boxed{\sf Mass= 110 g}}$

$\large{\sf In\:1L\:We\:Have\:110g}$

$\large{\boxed{\sf Mole(n)={\frac{Mass}{Molecular\:Weight}}}}$

We have calculated

Mass (m) = 110g

Molar mass of $\sf{H_{2}SO_{4}}$ = $\sf{ 2+32+16 \times 4}$

$\large{\sf Molecular\:Weight (M_{wt}) = 98}$

$\large{\sf n ={\frac{110}{98}}}$

$\large{\boxed{\sf n=1.12}}$

WE HAVE TO FIND MOLARITY :-

FORMULA TO FIND MOLARITY (M) IS :-

$\large{\boxed{\sf M ={\frac{Mole\:Of\:Solute}{Litre\:Of\:Solute}}}}$

$\large{\sf M = {\frac{1.12}{1}}}$

$\large{\boxed{\sf M = 1.12 Mol/litre}}$

Answer 2

Answer:

The molarity of aqueous solution of H₂SO₄ is equal to 1.12mol/L.

Explanation:

Given: the weight by weight concentration = 10%

Density of the solution $=1.1gml^{-1}$

Volume of solution $=1000ml$

We know, Mass = Density× Volume × Concentration

⇒ Mass $=(1.1)(1000)(0.1)=110g$

Molar mass of $H_{2}SO_{4}$ , $M_{wt}=2+32+64=98g$

Number of moles = Mass/Molar mass

Number of moles $=\frac{110}{98}=1.12mol$

Molarity = (moles of H₂SO₄ )/(Volume of solution in L)

So, molarity $=\frac{1.12}{1}=1.12molL^{-1}$

Therefore, The molarity of aqueous solution of sulphuric acid is 1.12molL⁻¹.