Find the purity percentage of $H_{2} {SO}_{4}$ with density $1.8$ , if $5 ml$ of $H_{2} {SO}_{4}$ is neutralized completely by $84.6 ml$ of $2 N$ $NaOH$ solution.

Process – A is a chemical change

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Process – B is a chemical change.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Both processes A and B are chemical changes.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these processes is a chemical change

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Titration:
Titration is the process of gradual addition of a solution of a known concentration and volume with another solution of unknown concentration until the reaction approaches its neutralization.
Titration is used to find the normality of the acid and base titration:
Molarity is No. of moles dissolved in $1000 ml$ solution.
Step-1 Calculating the Normality of $H_{2} {SO}_{4}$
N₁ $=$ Normality of the Acidic $H_{2} {SO}_{4}$ solution $=$ N
V₂ $=$ Volume of the Acidic $H_{2} {SO}_{4}$ solution $=$ $5 ml$
N₂ $=$ Normality of the basic $NaOH$ solution $=$ $2 N$
V₂ = Volume of the basic $NaOH$ solution $=$ $84.6 ml$
N₁V₁ $=$ N₂V₂
$N \times 5 = 84.6 \times 2$
$N = 33.84 N$
Step-2 Converting normality into molarity
Molarity $=$ Normality $\div$ basicity
$= 33.84 N \div 2 \Rightarrow 16.92 M$
$16.92$ moles of solutes dissolve in $1000 ml$ of solution to give $16.92 M$ solution
Step-3 Calculating the Mass of $H_{2} {SO}_{4}$ using density
Given the density of $H_{2} {SO}_{4}$ $=$ $1.8$
Volume of $H_{2} {SO}_{4}$ from molarity $=$ $1000 m L$
$Density = \frac{Mass}{Volume}$ ,
$Mass = Density \times Volume$
The mass of $H_{2} {SO}_{4}$ $=$ Number of moles of $H_{2} {SO}_{4}$ $\times$ Molecular weight $\Rightarrow$ $16.92 \times 98 \Rightarrow 1658.16 g$
Step-4 Calculating the Percentage purity of $H_{2} {SO}_{4}$
Percentage Purity $=$ (Mass of $16.92$ moles of $H_{2} {SO}_{4}$ $\div$ Mass of $H_{2} {SO}_{4}$ with density $1.8$ ) $\times 100$
$=$ $(1658.16 \div 1800) \times 100 \Rightarrow 92.12 %$
Hence, the purity percentage of $H_{2} {SO}_{4}$ with density $1.8$ = $92.12 %$

Suggest Corrections

Similar questions

Complete the table given below for writing the formula of a chemical compound.

Identify the valency

Criss cross the valency

Formula of the compound

Name of the compound

$Cr$	${SO}_{4}$

$Cr$	${SO}_{4}$

27. A non-metallic element is converted into a compound X after a series of reactions. A little amount of X when tested with blue litmus turns to red. X on complete reaction with another compound Y gave the product which did not respond to litmus test. Identify the correct sequence of the reactions.

(1) $2 {SO}_{2} + O_{2} \to 2 {SO}_{3}$

(2) $H_{2} {SO}_{4} + 2 KOH \to K_{2} {SO}_{4} + 2 H_{2} O$

$(3) {SO}_{3} + H_{2} O \to H_{2} {SO}_{4}$

$(4) S + O_{2} \to {SO}_{2}$

Q. 6.10 gram of impure naoh is completely neutralized by 1000 ml of 1/10n hcl .calculate the percentage purity of the impure naoh

Write the chemical names of the following compounds:

$K_{2} {SO}_{4}$

${Mg}_{3} {({PO}_{4})}_{2}$

${NH}_{4} Cl$

$ZnS$

${Na}_{3_{}} N$

$AgBr$

Balance the given chemical equation in a stepwise manner:

$HCl + Zn \underset{}{\to}$ ${ZnCl}_{2} + H_{2}$

Join BYJU'S Learning Program

Find the purity percentage of H2SO4with density1.8, if 5ml of H2SO4 is neutralized completely by 84.6mlof 2N NaOH solution.

Find the purity percentage of $H_{2} {SO}_{4}$ with density $1.8$ , if $5 ml$ of $H_{2} {SO}_{4}$ is neutralized completely by $84.6 ml$ of $2 N$ $NaOH$ solution.