Given density = 1.836 g/cm^3.
This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
7.a)
Step 1: Calculate the mass of $1000 \, \text{cm}^3$ of the concentrated acid.
Given density $= 1.836 \, \text{g/cm}^3$.
$$ \text{Mass} = \text{Density} \times \text{Volume} $$
$$ \text{Mass} = 1.836 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 1836 \, \text{g} $$
Step 2: Calculate the mass of pure sulphuric(VI) acid in $1000 \, \text{cm}^3$.
Given percentage purity $= 98\%$.
$$ \text{Mass of pure acid} = \frac{\text{Percentage purity}}{100} \times \text{Total mass} $$
$$ \text{Mass of pure acid} = \frac{98}{100} \times 1836 \, \text{g} = 0.98 \times 1836 \, \text{g} = 1799.28 \, \text{g} $$
Step 3: Calculate the moles of pure sulphuric(VI) acid.
Given relative formula mass $= 98$.
$$ \text{Moles} = \frac{\text{Mass}}{\text{Relative formula mass}} $$
$$ \text{Moles} = \frac{1799.28 \, \text{g}}{98 \, \text{g/mol}} = 18.36 \, \text{mol} $$
Step 4: Determine the concentration (molarity) of the acid.
Since the moles were calculated for $1000 \, \text{cm}^3$ ($1 \, \text{L}$) of solution, this is the molarity.
$$ \text{Concentration} = \frac{\text{Moles}}{\text{Volume in L}} = \frac{18.36 \, \text{mol}}{1 \, \text{L}} = 18.36 \, \text{M} $$
The concentration of the acid is $\boxed{\text{18.36 M}}$.
7.b)
Step 1: Use the dilution formula $M_1V_1 = M_2V_2$.
$M_1$ (concentration of concentrated acid) $= 18.36 \, \text{M}$ (from 7.a))
$V_1$ (volume of concentrated acid) $= ?$
$M_2$ (desired concentration of diluted acid) $= 2 \, \text{M}$
$V_2$ (desired volume of diluted acid) $= 2000 \, \text{cm}^3$
Step 2: Substitute the values into the formula and solve for $V_1$.
$$ M_1V_1 = M_2V_2 $$
$$ 18.36 \, \text{M} \times V_1 = 2 \, \text{M} \times 2000 \, \text{cm}^3 $$
$$ V_1 = \frac{2 \, \text{M} \times 2000 \, \text{cm}^3}{18.36 \, \text{M}} $$
$$ V_1 = \frac{4000}{18.36} \, \text{cm}^3 $$
$$ V_1 \approx 217.86 \, \text{cm}^3 $$
The volume of concentrated sulphuric(VI) acid needed is $\boxed{\text{217.86 cm}^3}$.
8.a)
The criteria for testing the purity of water include checking its boiling point and freezing point. Pure water boils at $100^\circ\text{C}$ and freezes at $0^\circ\text{C}$ at standard atmospheric pressure. Any deviation from these specific temperatures indicates the presence of impurities. Another criterion is electrical conductivity; pure water has very low conductivity, while impurities increase it.
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