What volume of 95% sulphuric acid (density = $1.85\,g/c{{m}^{3}}$) and what mass of water must be taken to prepare $100\,c{{m}^{3}}$of 15% solution of sulphuric acid (density = $1.10\,g/c{{m}^{3}}$) ?
Answer
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Hint: Density of any solution is the mass of that solution in per unit volume. So, density has a formula of mass upon volume. The percentage of any solution is the amount of the solution in 100 ml of the solution. For example 15% sulphuric acid contains 0.15 of the sulphuric acid dissolved in 100 ml of the solution.
Complete answer:
We have been given sulphuric acid solutions. One solution is 95 % and the other is 15%, along with their densities. We have to find two values, the volume of 95% sulphuric acid and mass of water to prepare $100\,c{{m}^{3}}$of 15% solution of sulphuric acid.
Taking out the mass of water, as we know density = $\dfrac{mass}{volume}$ , therefore, given is volume $100\,c{{m}^{3}}$and density $1.10\,g/c{{m}^{3}}$for water, we have
Mass = $V\times d$
Mass = $100\,c{{m}^{3}}\times $ $1.10\,g/c{{m}^{3}}$
Mass of 15% solution = 110 g
Now, 15% solution has mass = $0.15\times 110$ = 16.5
So, mass of water will be = 110 – 16.5 = 93.5 g
Now, calculating the volume of 95% sulphuric acid needed to prepare $100c{{m}^{3}}$ of 15% sulphuric acid solution.
Let 95% sulphuric acid solution require 16.5 g of water, and volume of acid in it be ‘x’. So,
16.5 g = 0.95$\times $ x
x = $16.5\times \dfrac{100}{95}$
x = 17.4 g
As, volume = $\dfrac{mass}{density}=\dfrac{17.4}{1.85}$
Volume = $9.4\,c{{m}^{3}}$
Hence, the volume of 95% sulphuric acid and mass of water needed to prepare 15% sulphuric acid solution is $9.4\,c{{m}^{3}}$ and 93.5 g respectively.
Note:
From the volume of 95% sulphuric acid that is $9.4\,c{{m}^{3}}$ we can get volume of water as 100 – 9.4 = $90.6\,c{{m}^{3}}$. As density gets affected by the mass and volume, the change in the concentration of the sulphuric acid as 15% and 95% changes the density of sulphuric acid as $1.10\,g/c{{m}^{3}}$ and $1.85\,g/c{{m}^{3}}$respectively.
Complete answer:
We have been given sulphuric acid solutions. One solution is 95 % and the other is 15%, along with their densities. We have to find two values, the volume of 95% sulphuric acid and mass of water to prepare $100\,c{{m}^{3}}$of 15% solution of sulphuric acid.
Taking out the mass of water, as we know density = $\dfrac{mass}{volume}$ , therefore, given is volume $100\,c{{m}^{3}}$and density $1.10\,g/c{{m}^{3}}$for water, we have
Mass = $V\times d$
Mass = $100\,c{{m}^{3}}\times $ $1.10\,g/c{{m}^{3}}$
Mass of 15% solution = 110 g
Now, 15% solution has mass = $0.15\times 110$ = 16.5
So, mass of water will be = 110 – 16.5 = 93.5 g
Now, calculating the volume of 95% sulphuric acid needed to prepare $100c{{m}^{3}}$ of 15% sulphuric acid solution.
Let 95% sulphuric acid solution require 16.5 g of water, and volume of acid in it be ‘x’. So,
16.5 g = 0.95$\times $ x
x = $16.5\times \dfrac{100}{95}$
x = 17.4 g
As, volume = $\dfrac{mass}{density}=\dfrac{17.4}{1.85}$
Volume = $9.4\,c{{m}^{3}}$
Hence, the volume of 95% sulphuric acid and mass of water needed to prepare 15% sulphuric acid solution is $9.4\,c{{m}^{3}}$ and 93.5 g respectively.
Note:
From the volume of 95% sulphuric acid that is $9.4\,c{{m}^{3}}$ we can get volume of water as 100 – 9.4 = $90.6\,c{{m}^{3}}$. As density gets affected by the mass and volume, the change in the concentration of the sulphuric acid as 15% and 95% changes the density of sulphuric acid as $1.10\,g/c{{m}^{3}}$ and $1.85\,g/c{{m}^{3}}$respectively.
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