Question

Calculate the mass percent (w/w) of sulphuric acid in a solution prepared by dissolving 4g of sulphur trioxide in a 100 mL sulphuric acid solution containing 80 mass percent (w/w) of $$H_{2}S{O}_4$$ and having a density of 1.96 g/ml(molecular weight of $$H_{2}S{O}_4$$ = 98) Take reaction
$$S{O}_3+H_{2}O\to H_{2}S{O}_4$$
A. 80.8%
B. 84%
C. 41.65%
D. None of these

Answer 1

Hint: The mass percentage of the solution is calculated by dividing mass of solute by total mass of solution multiplied by 100. The mass of sulphuric acid is calculated by the formula of density where mass is divided by the volume.

Complete step by step answer:
Given,
Mass of the Sulphur trioxide is 4g.
Volume of sulphuric acid is 100mL.
Mass percent of sulphuric acid is 80%
The density is 1.96g/mL
The molecular weight of sulphuric acid is 98.
The formula of calculating the density is shown below.
$$density={\displaystyle \frac{mass}{volume}}$$
To calculate the mass of the sulphuric acid solution, substitute the values in the above equation.
$$\Rightarrow 1.96g/mL={\displaystyle \frac{mass}{100mL}}$$
$$\Rightarrow mass=1.96g/mL\times 100mL$$
$$\Rightarrow mass=196g$$
$$\Rightarrow$$The mass of sulphuric acid = $${\displaystyle \frac{80}{100}}\times 196$$
$$\Rightarrow$$The mass of sulphuric acid = 156.8 g
Weight of water = weight of solution – weight of sulphuric acid.
$$\Rightarrow$$Weight of water = 196 – 156.8
$$\Rightarrow$$Weight of water = 39.2g
The reaction is shown below.
$$S{O}_3+H_{2}O\to H_{2}S{O}_4$$
80 g Sulphur trioxide reacts with 18 g of water to form 98 g sulphuric acid.
As 4g of Sulphur trioxide is added so mass of water is calculated as shown below.
$$\Rightarrow {\displaystyle \frac{18}{80}}\times 4$$
$$\Rightarrow 0.9g$$
The mass of sulphuric acid is calculated as shown below.
$$\Rightarrow {\displaystyle \frac{98}{80}}\times 4$$
$$\Rightarrow 4.9g$$
The final weight of water = 39.2 – 0.9
$$\Rightarrow$$ The final weight of water = 38.3
The final weight of sulphuric acid = 156.8 + 4.9
$$\Rightarrow$$The final weight of sulphuric acid = 161.7
The formula to calculate the mass percent is shown below.
$$mass\mathrm{\%}={\displaystyle \frac{massofsolute}{massofsolution}}\times 100$$
To calculate the mass percent substitute the values in the above equation.
$$\Rightarrow mass\mathrm{\%}={\displaystyle \frac{161.7}{161.7+38.3}}\times 100$$
$$\Rightarrow mass\mathrm{\%}={\displaystyle \frac{161.7}{200}}\times 100$$
$$\Rightarrow mass\mathrm{\%}=80.85\mathrm{\%}$$
Thus, the mass percent of sulphuric acid solution is 80.85 %.

So, the correct answer is Option A.

Note: The mass of one mole of compound is equal to the molecular weight of the compound. During the reaction water is reduced therefore the mass is deducted to get the final weight of the water.