Question

44. Aqueous solution of H2SO4 contains 68% H2SO4 by
mass. Its density is 1.12 g ml-!. What will be its molarity
(M) and molality (m)?
(a) M = 7.77, m = 21.68 (b) M= 21.68, m= 7.77
(c) M= 7.77, m= 7.77 (d) M = 21.68, m= 21.68
​

Answer 1

Answer:

ANSWER

98% by weight of solution means 98g of H2SO4 is present in 100g of solution.

Volume of solution =density of solutionmass of solution=(1.8×1000)g/L100g

Volume of solution =181L

Per cent  H2SO4 w/v of solution = molarity =volume of solution (in L)number of moles of H2SO4

Molarity =volume of solutiongiven mass of H2SO4/ molar mass of H2SO4

Molarity =1/1898/98=18M

Answer 2

Given:

Aqueous solution of sulphuric acid containing $\[68\% \,{H_2}S{O_4}\]$ by mass

Density of $\[{H_2}S{O_4} = 1.12\,g\,m{l^{ - 1}}\]$

To Find: Molarity (M) and molality (m)

Solution:

Since the solution contains $\[68\% \,{H_2}S{O_4}\]$ by mass, therefore, it means it has 68 g of $\[{H_2}S{O_4}\]$ is present in 100 g of solution.

Therefore, the volume of the solution can be given as:

$\[Volume\,of\,solution = \frac{{Mass}}{{Density}}\]$

$\[V = \frac{{100\,g}}{{1.12\,g\,m{l^{ - 1}}}}\]$

$\[V = 89.28\,ml\]$

Number of moles of $\[{H_2}S{O_4}\]$ present in the solution can be given as:

$\[n = \frac{{mass}}{{molar\,mass}}\]$

$\[n = \frac{{68\,g}}{{98\,g\,mo{l^{ - 1}}}}\]$

$\[n = 0.6938\,mol\]$

Therefore, the molarity (M) of the solution is given as:

$\[M = \frac{n}{V} \times 1000\]$

$\[M = \frac{{0.6938}}{{89.28}} \times 1000\]$

$\[M = 7.77\,mol/l\]$

Molality of the solution can be given as

$\[m = \frac{n}{w} \times 1000\]$

$\[m = \frac{{0.6938}}{{32\,g}} \times 1000\]$

$\[m = 21.68\,mol/kg\]$

Hence, the molarity (M) and molality (m) of the solution are $\[7.77\,mol/l\]$ and $\[21.68\,mol/kg\]$ respectively. Thus, the correct option is (a).