Question

A sample of sulphuric acid contains 13% H2SO4 by mass and its density is 1.02 g/ml. Calculate the molality and morality of the sample. ​

Answer 1

Answer:

The molality and molarity of the sample are 1.52 mol/kg ans 1.35 mol/l respectively.

Explanation:

13% of sulfuric acid solution means that in 100 grams of solution  there are 13 g of sulfuric acid and 87 g of water..

$m=\frac{13 g}{98 g/mol\times 0.087 kg}=1.52 mol/kg$

Density of thye solution = d= 1.02 g/ml

Volume of the solution:

$V=\frac{mass}{density}=\frac{100 g}{1.02 g/ml}=98.03 ml$

$Molarity=\frac{13 g}{98 g/mol\times 0.098 L}=1.35 mol/L$

The molality and molarity of the sample are 1.52 mol/kg ans 1.35 mol/l respectively.

Answer 2

Answer:

Molarity = 1.3525 mole/liter

Molality = 1.5241 mole/Kg

Explanation:

Let us consider 100 grams of the solution.

density of solution = 1.02 g/ml

Volume of solution = mass/density

                                = 100/1.02

                                = 98.04 ml

                                = 0.09804 L

Mass percent  = 13%

Hence, in 100 grams of the solution, we have

mass of solute (sulphuric acid) = 13 g

Mass of solvent (water) = 87 g

Molar mass of sulphuric acid = 98 g/mol

⇒Number of moles in the sample = 13/98 = 0.1326 moles

CALCULATION OF MOLARITY

Molarity is defined as number of moles of solute present in 1 liter of the solution.

Here we have

Number of moles = 0.1326 moles

Volume of solution = 0.09804 L

Hence, Molarity = Number of moles/Volume of solution in liter

                            = 0.1326/0.09804

                            = 1.3525 M

CALCULATION OF MOLALITY

Molality is defined as the number of moles of solute present in 1 kg of the solvent

Here we have

Number of moles of solute  = 0.1326 moles

Mass of solvent = 87 g = 0.087 Kg

Molality = Number of moles/Mass of solvent

             = 0.1326/0.087

             =  1.5241