The wavenumber ($\tilde{\nu}$) is given by:
\[ \tilde{\nu} = \frac{1}{\lambda}, \]
where $\lambda$ is the wavelength in cm.
Step 1: Convert wavelength to cm
\[ \lambda = 5800 \, \text{\AA} = 5800 \times 10^{-8} \, \text{cm}. \]
Step 2: Calculate wavenumber
\[ \tilde{\nu} = \frac{1}{5800 \times 10^{-8}} = \frac{1}{5.8 \times 10^{-5}} = 17241 \, \text{cm}^{-1}. \]
Step 3: Express as $x \times 10 \, \text{cm}^{-1}$
\[ 17241 \, \text{cm}^{-1} = 1724 \times 10 \, \text{cm}^{-1}. \]
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)