0.4 g mixture of NaOH, Na₂CO₃ and some inert impurities was first titrated with $\frac{N}{10}$ HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After his methyl orange was added and titrated. 1.5 mL of same HCl was required for the next end point. The weight percentage of Na₂CO₃ in the mixture is- (Rounded-off to the nearest integer)

1. 6 %
2. 4 %
3. 7 %
4. 9 %

Complex A has a composition of H₁₂O₆Cl₃Cr . If the complex on treatment with concentrated H₂SO₄ loses 13.5% of its original mass, the correct molecular formula of A is : 
[ Given: atomic mass of Cr = 52 amu  and Cl =35 amu]

1. [Cr(H₂O)₅Cl]Cl₂$·$H₂O
2. [Cr(H₂O)₆]Cl₃
3. [Cr(H₂O)₃Cl₃]$·$3H₂O
4. [Cr(H₂O)₄Cl₂]Cl$·$2H₂O

A 100 mL solution was made by adding 1.43 g of Na₂CO₃.xH₂O. The normality of the solution is 0.1 M. The value of x is: 
(The atomic mass of Na is 23 g/mol)

1. 1 
2. 10 
3. 4
4. 7

The volume, in mL, of 0.02 M K₂Cr₂O₇ solution required to react with 0.288 g of ferrous oxalate in an acidic medium is: 
Molar mass of Fe = 56 g mol^–1)

1. 55 mL
2. 45 mL
3. 50 mL
4. 60 mL

A solution of two components containing ${\mathrm{n}}_1$ moles of the 1^st component ${\mathrm{n}}_2$ moles of the 2^nd component is prepared. ${\mathrm{M}}_1 \mathrm{and} {\mathrm{M}}_2$ are the molecular weights of component 1 and 2 respectively. If d is the density of the solution in g ${\mathrm{mL}}^{-1}$, ${\mathrm{C}}_2$ is the molarity and ${\mathrm{x}}_2$ is the mole fraction of the 2^nd component, then ${\mathrm{C}}_2$ can be expressed as: 

1. ${\mathrm{C}}_2=\frac{{\mathrm{dx}}_2}{{\mathrm{M}}_2+{\mathrm{x}}_2\left({\mathrm{M}}_2-{\mathrm{M}}_1\right)}$

2. ${\mathrm{C}}_2=\frac{1000 {\mathrm{dx}}_2}{{\mathrm{M}}_1+{\mathrm{x}}_2\left({\mathrm{M}}_2-{\mathrm{M}}_1\right)}$

3. ${\mathrm{C}}_2=\frac{1000{\mathrm{x}}_2}{{\mathrm{M}}_1+{\mathrm{x}}_2\left({\mathrm{M}}_2-{\mathrm{M}}_1\right)}$

4. ${\mathrm{C}}_2=\frac{{\mathrm{dx}}_1}{{\mathrm{M}}_2+{\mathrm{x}}_2\left({\mathrm{M}}_2-{\mathrm{M}}_1\right)}$

A solution of phenol in chloroform when treated with aqueous NaOH gives compound P as a major product. The mass percentage of carbon in P is:
(to the nearest integer) (Atomic mass : C =12; H=1; O=16)