Question

The pH of a sample of $ {{H}_{2}}S{{O}_{4}} $ is $ 1.3979. $ The percentage of the solution is $ 73.5%\text{ }\left( w/w \right), $ the density of the solution is:
(A) $ 2.66\times {{10}^{-3}}\text{ }g/cc $
(B) $ 5.32\times {{10}^{-3}}\text{ }g/cc $
(C) $ 1.33\times {{10}^{-3}}\text{ }g/cc $
(D) $ 0.01\text{ }g/cc $

Answer 1

Hint :We know that the reaction of an acid and a base to form salt and water is called neutralization. First of all we shall calculate the milliequivalents of sulphuric acid and then its mass can be calculated. Then using the percentage composition, the density mass of the solution can be calculated.

Complete Step By Step Answer:
Molarity and normality are units of concentration in chemistry. Molarity is defined as the number of moles of a solute dissolved in one litre of a solution. Normality: Normality is defined as the number of equivalents per litre of solution. In other words, it is a measure of concentration that is equal to gram equivalent weight of solute per litre of solution. The mass percentage of the solution is calculated by dividing the mass of the solute by total mass of solution multiplied by $ 100. $ The mass of sulphuric acid is calculated by the formula of density where mass is divided by the volume. N-factor basically defines the amount of the reacting species. The n factor varies as per the reaction and nature of the substance. It is also defined as the valency of the ions in the aqueous solution.
Given that, $ pH=1.3979 $ also we know that $ \left[ {{H}^{+}} \right]=0.04M $
 $ {{H}_{2}}S{{O}_{4}}_{\left( aq \right)}\to 2H_{\left( aq \right)}^{+}+SO_{4}^{2-} $
Since, $ \left[ {{H}^{+}} \right]=0.04M, $ hence $ \left[ {{H}_{2}}S{{O}_{4}} \right]=0.02M $ which is One Litre of solution has $ 0.02 $ mole $ {{H}_{2}}S{{O}_{4}} $
Let the density be $ D\text{ }gm/mL $ then mass of solution $ =1000\text{ }D $
 $ %\text{ }w/w\text{ }=\dfrac{0.02\times 98}{1000\text{ }D}=\dfrac{73.5}{100}. $
From here we get $ D=\dfrac{0.02\times 98}{10\times 73.5}\text{ }gm/mL $
On further solving we get; $ D=2.66\times {{10}^{-3}}\text{ }gm/cc $ .
Therefore, the correct answer is option A.

Note :
Remember that the mass of one mole of compound is equal to the molecular weight of the compound. During the reaction water is reduced therefore the mass is deducted to get the final weight of the water.