WHAT VOLUME OF 95 % SULPHURIC ACID ( DENSITY = 1.85 g cm^-3 ) AND WHAT MASS OF WATER MUST BE TAKEN TO PREPARE 100 cm^3 OF 15 % SOLUTION OF SULPHURIC ACID ( DENSTIY = 1.10 g cm^-3)?

Asked by kandappan | 2nd Apr, 2020, 08:07: PM

Expert Answer:

To find:
Volume of 95% sulphuric acid (density = 1.85 g/mL)
Mass of water to prepare 100 mL of 15% solution of sulphuric acid (density = 1.10 g/mL)
 
Solution:

Volume of the solution = 100 cm2

Density of the solution = 1.10 g/cm3

Therefore, Mass of 100 cm3 of solution = 100 x 1.10 g =110 g

The given solutions is 15%. Therefore, 100 g of solution contains 15 g of H2SO4

Then, mass H2SO4 in 110 g (100 cm3 ) of solution = begin mathsize 11px style fraction numerator 15 space straight g over denominator 100 space straight g end fraction space cross times space 110 space straight g space equals bold space bold 16 bold. bold 5 bold space bold g end style

Mass of water in 110 g (100cm3 ) of solution = (110 - 16.5) g = 93.5 g

To obtain 100 cm3 of 15 % solution acid, we require the given information.

Mass of water = 93.5 g

Mass of H2SO(100% pure) = 16.5 g

Since, the given sulphuric acid is 95% pure, hence,

Mass of H2SO4 (95%) will be required =

begin mathsize 11px style fraction numerator 100 space cross times space 16.5 space straight g over denominator 95 end fraction space equals space 17.37 space straight g Density space of space 95 % space straight H 2 SO 4 space equals space 1.85 space cm to the power of negative 3 end exponent end style

Answered by Ramandeep | 3rd Apr, 2020, 11:03: AM

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