At 25°C, an aqueous solution containing 35.0 wt% H2SO4 has a specific gravity of 1.2563. A quantity of the 35% solution is needed that contains 195.5 kg of H2SO4. A) Calculate the required volume (L) of the solution using the given specific gravity.
B) Estimate the percentage error that would have resulted if pure-component specific gravities of H2SO4 (SG = 1:8255) and water had been used for the calculation instead of the given specific gravity of the mixture.

a) volume₁ = 444.6 L  

b) Volume₂ = 306 L  and percentage Error = 31.2%

Explanation:      

Given that;

the solution contains 35.0 wt% H₂SO₄

A quantity of the 35% solution is needed that contains 195.5 kg of H₂SO₄

Lets say mass of solution containing 195.5 kg H₂SO₄ is 'A' kg

Now since the question saysm it is a 35% wt solution,

so

(35/ 100) × Akg = 195.5kg

0.35A = 195.5

A = 558.6kg

So A = 558.6 kg

therefore mass of the  solution is 558.6kg

a)

also Specific gravity is 1.2563

since density of water = 1kg/ L

density of solution = SG of H₂SO₄ × density of water

therefore density of solution = 1.2563 ×1kg/ L  = 1.2563 kg/ L

Now to calculate the required volume (L) of the solution

we say;

Volume of solution = mass / density

Volume = 558.6kg / 1.2563kg/L

Volume₁ = 444.6 L  

b)

Now If pure-component specific gravity is to be used,

Specific Gravity = 1.8255

which means Density will be  = 1.8255 kg/ L

Therefore will be

Volume = 558.6kg / 1.8255kg/L

Volume₂ = 306 L

To calculate the error

we say volume₁ - volume₂

Error = 444.6L - 306L = 138.6

So

Percent error = ( 138.6L / 444.6L) × 100

percentage Error = 31.2%

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.