Question

Calculate the molality of 9.8 parcent ( w/ w) solution of sulphuric acid given the density of solution is 1.02 g/ ml

Answer 1

The molality of the solution is 1.02 M.

Accurate Question :

Calculate Molality of 9.8% (w/w) solution of sulphuric acid $\bf H_2SO_4$ . If the density of the solution is 1.02 g/ml.

Explanation :

Given :

Density of the solution - 1.02 g/ml.

Molar mass of   $\bf H_2SO_4$ - 98 g/ mol.

Mass of  $\bf H_2SO_4$ - 9.8.

Mass of solution - 100 grams.

9.8% (w/w) solution of sulphuric acid.

This implies that -

9.8 grams of $\bf H_2SO_4$ is present in 100 gram of solution.

To Find :

The Molality.

We know that :

$\bigstar\;\boxed{\textbf{Molality of the solution}=\dfrac{\textbf{Moles of the solute}}{\textbf{Volume of the solution (in L)}}}}}\\ \\ \\\bigstar\;\boxed{\textbf{Density}=\dfrac{\bf Mass}{\bf {Volume}}}}}}}}}\\ \\ \\\bigstar\;\boxed{\textbf{Number of moles}=\dfrac{\textbf{Given mass}}{\textbf{Molar\;mass}}}$

So,

The volume of the solution :

$\implies \sf{1.02g/mL=\dfrac{100g}{\textsf{Volume of the solution}}}}\\ \\ \\ \textsf{Volume of the solution}=98.03mL\\ \\ \\\implies 0.098L$

The number of moles :

$\implies \sf \dfrac{9.8g}{98g/mol}\\ \\ \\\implies \sf 0.1mol$

Putting values,

$\implies \textsf{Molality of the solution}=\dfrac{\sf 0.1mol}{\sf 0.098L}\\ \\ \\\implies \sf 1.02mol/L$

Answer 2

Answer:

1.02 mol/l

this is answer