Answer
Verified
379.2k+ views
Hint: We need to realize that oxalic corrosive is having the equation ${H_2}{C_2}{O_4}$ . It is likewise called crab corrosive. It is really a white glasslike strong which is by and large found to get broken up in arrangement and colourless. The equation used to ascertain the load for oxalic corrosive must be given underneath,
${N_1}{V_1} = \dfrac{{Weight \times Valencyfactor}}{{Molarmass}} \times 1000$
Here,
${N_1}$ = Normality of $MnO_4^ - $
${V_1}$ = Volume of $MnO_4^ - $
Complete answer:
We need to know, the arrangement that is the $100mL$ of $MnO_4^ - $ , at that point it is inquired as to whether we add $10mL$ in $50mL$ of $10N$ ${I^ - }$ , at that point it will be completely changed over into , so we can compose its condition,
$M{n^{ + 7}} + 5{e^ - } \to M{n^{ + 2}}$
Then,
${I^ - } + {e^ - } \to {I_2}$
We can write the expression as,
${N_1}{V_1} = {N_2}{V_2}$\[\]
Here,
${N_1}$ = Normality of $MnO_4^ - $
${V_1}$ = Volume of $MnO_4^ - $
${N_2}$ = Normality of ${I^ - }$
${V_2}$ = Volume of ${I^ - }$
To calculate the normality ${N_1}$ ,
Applying given values in the above expression,
${N_1} \times 10 = 50 \times 10$
Then,
${N_1} = 5$
Now, we have to calculate mass of the ${H_2}{C_2}{O_4}$ by the use of the value of molar mass of ${H_2}{C_2}{O_4}$ .The molar mass of ${H_2}{C_2}{O_4}$ = $90g/mol$ .
Where, we can see that carbon in oxalic corrosive in $ + 3$ state, and it is evolving to, so the valency of this is $2$ , and reciprocals are something similar.
${N_1}{V_1} = \dfrac{{Weight \times Valencyfactor}}{{Molarmass}} \times 1000$
Applying given values in the above equation,
$5 \times 10 = \dfrac{{w \times 2}}{{90}} \times 100$
Then, to find out the weight,
$w = \dfrac{{5 \times 10 \times 90}}{{2 \times 100}} = \dfrac{{4500}}{{200}} = 22.5$
Therefore,
The weight is $22.5g$
Hence, the option (D) is correct.
Note:
We need to establish ordinariness too. In this way, we ought not to get confused regarding ordinariness and molarity. At the point when ordinariness is the quantity of the same solute disintegrated per liter of arrangement. $N$ is represented as normality, and $M$ is represented as molarity.
${N_1}{V_1} = \dfrac{{Weight \times Valencyfactor}}{{Molarmass}} \times 1000$
Here,
${N_1}$ = Normality of $MnO_4^ - $
${V_1}$ = Volume of $MnO_4^ - $
Complete answer:
We need to know, the arrangement that is the $100mL$ of $MnO_4^ - $ , at that point it is inquired as to whether we add $10mL$ in $50mL$ of $10N$ ${I^ - }$ , at that point it will be completely changed over into , so we can compose its condition,
$M{n^{ + 7}} + 5{e^ - } \to M{n^{ + 2}}$
Then,
${I^ - } + {e^ - } \to {I_2}$
We can write the expression as,
${N_1}{V_1} = {N_2}{V_2}$\[\]
Here,
${N_1}$ = Normality of $MnO_4^ - $
${V_1}$ = Volume of $MnO_4^ - $
${N_2}$ = Normality of ${I^ - }$
${V_2}$ = Volume of ${I^ - }$
To calculate the normality ${N_1}$ ,
Applying given values in the above expression,
${N_1} \times 10 = 50 \times 10$
Then,
${N_1} = 5$
Now, we have to calculate mass of the ${H_2}{C_2}{O_4}$ by the use of the value of molar mass of ${H_2}{C_2}{O_4}$ .The molar mass of ${H_2}{C_2}{O_4}$ = $90g/mol$ .
Where, we can see that carbon in oxalic corrosive in $ + 3$ state, and it is evolving to, so the valency of this is $2$ , and reciprocals are something similar.
${N_1}{V_1} = \dfrac{{Weight \times Valencyfactor}}{{Molarmass}} \times 1000$
Applying given values in the above equation,
$5 \times 10 = \dfrac{{w \times 2}}{{90}} \times 100$
Then, to find out the weight,
$w = \dfrac{{5 \times 10 \times 90}}{{2 \times 100}} = \dfrac{{4500}}{{200}} = 22.5$
Therefore,
The weight is $22.5g$
Hence, the option (D) is correct.
Note:
We need to establish ordinariness too. In this way, we ought not to get confused regarding ordinariness and molarity. At the point when ordinariness is the quantity of the same solute disintegrated per liter of arrangement. $N$ is represented as normality, and $M$ is represented as molarity.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE