Question

Calculate molarety and molality of sulphuric acid solution
of density 1.198 gm cm^-3 containing 27% by mass of
sulphuric acid ( Molar mass of H2SO4= 98 gm)​

Answer 1

Answer:

Molarity of the solution: 3.3007 mol/L

Molality of the solution: 3.7741 mol/kg

Explanation:

Percentage of the sulfuric acid = 27%

In 100 g of solution there is 27 g of sulfuric acid.

Mass of the sulfuric acid= 27 g

Mass of the solution = 100 g

Mass of solvent = 100 g - 27 g = 73 g = 0.073 kg

Density of the Solution = d = $1.198 g/ cm^3$

Volume of the solution = V

$density=\frac{mass}{volume}$

$V=\frac{100 g}{1.198 g/cm^3}=83.47 cm^3=83.47 mL=0.08347 L$

Moles of sulfuric acid =$\frac{27 g}{98 g/mol}=0.2755 mol$

Molarity of the solution:

$M=\frac{\text{Moles of}H_2SO_4}{\text{Volume of the solution (L)}}$

$M=\frac{27 g}{98 g/mol\times 0.08347 L}=3.3007 mol/L$

Molality of the solution:

$m=\frac{\text{Moles of}H_2SO_4}{\text{Mass of the solvent (kg)}}$

$m=\frac{27 g}{98 g/mol\times 0.073 kg}=3.7741 mol/kg$

Answer 2

Answer:

molarity = 3.77 , Volume = 83.47 ×10^-3

molality = 3.301

Explanation:

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