Some Basic Concepts of Chemistry

Number of sigma bonds are 10.
[Structure showing 10 sigma bonds in the molecule]

Number of significant figure in 0.00340 are 3.

W = 4.5 g

M.W = 90

V = 250 ml

Using M = $\frac{W}{M.W*V}*1000$  

$M=\frac{4.5}{90*250}*1000=0.2$  

M = 2 * 10^-1 M

So, x = 2

Mass of Na⁺ in 50 ml = 70 * 50 = 3500 mg

23000 mg of Na⁺ is present in 85000 mg of NaNO₃ (1 mole NaNO₃ contains 1 mole Na⁺)

$\therefore 3500$ mg Na⁺ will be present in   $\frac{85000}{23000}*3500$

= 12934.78 mg

= 12.93478 gm

$\cong 13$        

M_eq of NaOH = M_eq H₂C₂O₄

4.44 * N = 1.25 * 2 * 10

$\therefore N=\frac{1.25*2*10}{4.44}=5.63$

$\therefore N=M=5.63$