Question

concentrated sulphuric acid has density 1.9g/ml and is 99% h2so4 by weight calculate the molality of the sulphuric acid in the acid ​

Answer 1

Molar mass of sulfuric acid=98g/mole

Mass of sulfuric acid in 100 g of solution = 99g

Mass of water in 100 g of solution =1g = 0.001 kg

Number of moles of sulfuric acid in 100 g of solution = 99/98 =1.01 moles

Molality = number of moles of solute/ mass of solvent in kg

= 1.01/0.001

=1010 m

Answer 2

Answer: The molarity of the solution comes out to be 19.19 M.

Explanation:

We are given:

99% m/m $H_2SO_4$. This means that 99 grams of sulfuric acid is present in 100 grams of solution.

Density of solution = 1.9 g/mL

To calculate the volume of the solution, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of solution = 1.9 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

$1.9g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=52.63mL$

To calculate the molarity of a solution, we use the equation:

$\text{Molarity}=\frac{\text{Number of moles of solute}}{\text{Volume of solution (in L)}}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of sulfuric acid = 99 g

Molar mass of sulfuric acid = 98g/mol

Volume of solution = 52.63 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{99\times 1000}{98\times 52.63}=19.19M$

Hence, the molarity of the solution comes out to be 19.19 M.