## What is n factor of K2Cr2O7 in basic medium?

Answer: N-factor for in basic medium is 3.

What is the equivalent weight of K2Cr2O7 in acidic medium?

Hence, the equivalent weight of \${K_2}C{r_2}{O_7}\$ in acidic medium will be 49g/eq.

What is the Colour of K2Cr2O7 in basic medium?

yellow
In acidic medium, K2Cr2O7 exists as Cr2O7^2 – (orange) while in basic medium it is converted to CrO4^2 – (yellow).

### How do you find the N factor of K2Cr2O7 in acidic medium?

oxidation state of Cr in K2Cr2O7 is +6….

1. Molecular weight of K2Cr2O7= 294.2.
2. In acidic medium 1- mole of K2Cr2O7 liberates 3- atoms of oxygen as follows.
3. K2CR2O7+ 4 H2SO4 ====> K2SO4 + Cr2(SO4)3+ 4 H2O + 3(O).
4. {Equivalent weight of an( OXIDISING AGENT) Compound is the weight of the Compound that liberates 8- g of oxygen.}

Why K2Cr2O7 is orange in acidic and yellow in basic medium?

Answer: This is because in basic medium, potassium dichromate forms chromate ions which is yellow in colour while in acidic medium forms dichromate ions. Dichro. ate ions are red in colour while chromateions are yellow in colour.

What is n factor of KMnO4 in neutral medium?

In neutral medium KMnO4 forms MnO2 (+7 to +4) i.e. it gains 3 electrons. So, n-factor= 3.

#### What is the equivalent weight of FAS?

-Hence, the equivalent weight of ferrous ammonium sulphate is 392.13g/mol.

How do you find the equivalent weight of an acid medium?

1. Answer: (4) Equivalent weight can be calculated by using the formula. Equivalent weight = Molar mass/number of electrons lost or gained (n)
2. In acidic medium: MnO4- + 8H +5e– –> Mn2 + 4H2O. Here n = 5.
3. In basic medium: MnO4−+ e− → MnO42− Here n = 1.
4. In neutral medium: Mno4– + 4H+ + 3e− –> MnO2+ 2H2O. Here n = 3.

Can we use dichromate in basic medium?

Potassium dichromate acts as an oxidising agent only in acidic medium. It doesn’t act as oxidising agent in basic medium because as it involves in non redox reaction.

## Why is K2Cr2O7 an oxidizing agent in acidic medium?

In acidic medium, Cr undergoes a decrease in oxidation state from +6 in Cr2O2-7 ion to +3 in Cr3+ ion. Since +3 oxidation state is more stable than +6 oxidation state therefore, K2Cr2O7 is a good oxidising agent in the acidic medium.

How do you find the N factor of Cu2S?

So, n-factor of Cu2S=4−(−2)+2×(2−1)=2+6=8.

Why K2Cr2O7 show different colour in acidic and basic medium?

It’s colour changes from orange to yellow due to formation of chromate (cro4-2) while in basic med. It again changes from yellow to orange.

### What is the equivalent weight of K2Cr2O7 in g mol?

The equivalent weight of K 2 Cr 2 O 7 = 49 g/mol In the solution, K 2 Cr 2 O 7 gives the following ions, K 2 Cr 2 O 7 + 14H+ + 6e – → 2K + + 2Cr 3+ + 7H 2 O The chemical equivalent is 6 because six electrons participate in the chemical reaction.

What is the equivalent weight of N A H 2 C2O4?

Therefore the equivalent weight is = 6M . Was this answer helpful? N a H C 2 ​ O 4 ​ is neutralised by N a O H and can also be oxidised by K M n O 4 ​ (in acidic medium). Equivalent weight is related to molecular weight ( M) of N a H 2 ​ C 2 ​ O 4 ​ in these reactions as :

What is the equivalent weight of potassium dichromate in acidic solution?

Potassium dichromate in acidic solution results in There is overall gain of 6 electrons. Therefore the equivalent weight is = 6M . Was this answer helpful? N a H C 2 ​ O 4 ​ is neutralised by N a O H and can also be oxidised by K M n O 4 ​ (in acidic medium).

#### What is the x/n-factor of K2Cr2O7 in acidic medium?

Hi, I hope my answer helps you;The x/n – factor of k2cr2o7 In acidic medium is 6 and in basic medium is 3. Provide a better Answer & Earn Cool Goodies See our forum point policy