Elementary Question on Relative Density

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SUMMARY

The discussion centers on the calculation of volume contraction when mixing 50g of sulphuric acid with a relative density of 1.84 and 50g of water. The resulting mixture has a relative density of 1.40, leading to a calculated contraction in volume of 28.571 cm³ per 100 cm³ of the mixture. The density of the acid is confirmed as 1.84 g/cm³, and the density of the mixture is established at 1.40 g/cm³. The initial volume of the dry acid is questioned, indicating it is not simply 50 cm³.

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  • Basic knowledge of volume and density relationships
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John O' Meara
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50g of sulphuric acid of relative density 1.84 are mixed with 50g of water, and the relative density of the mixture is found to be 1.40. Calculate the contraction in volume which has occurred? My attempt follows.

R.D., = density of acid/density of water => density of acid =1.84g/cm^3.
A R.D., of 1.40 => a density rho = 1.40g/cm^3.

100g divided by 1.40g/cm^3 = 71.429cm^3.

The contraction in the volume is 28.571cm^3 per 100cm^3 of the mixture!
 
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John O' Meara said:
50g of sulphuric acid of relative density 1.84 are mixed with 50g of water, and the relative density of the mixture is found to be 1.40. Calculate the contraction in volume which has occurred? My attempt follows.

R.D., = density of acid/density of water => density of acid =1.84g/cm^3.
A R.D., of 1.40 => a density rho = 1.40g/cm^3.

100g divided by 1.40g/cm^3 = 71.429cm^3.

The contraction in the volume is 28.571cm^3 per 100cm^3 of the mixture!

What's the initial volume of the dry acid? (It's not 50 cm^3).
 

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