0.1 mol of the following given antiviral compound (P) will weigh .........x $ 10^{-1} $ g.
Molar mass is given as 372 g/mol for compound (P).
Hence, for 0.1 mole, the mass will be: \[ \text{Mass} = \text{Molar mass} \times \text{Number of moles} = 372 \times 0.1 = 37.2 \, \text{g} \]
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: