Elementary Question on Relative Density

AI Thread Summary
The discussion focuses on calculating the contraction in volume when 50g of sulphuric acid with a relative density of 1.84 is mixed with 50g of water. The density of the acid is determined to be 1.84g/cm^3, while the mixture has a relative density of 1.40, equating to a density of 1.40g/cm^3. The total volume of the mixture is calculated to be 71.429cm^3 for 100g, indicating a contraction of 28.571cm^3. There is also a query regarding the initial volume of the dry acid, suggesting it is not simply 50cm^3. The conversation emphasizes the importance of understanding density and volume changes in mixtures.
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 occured? 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 occured? 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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