At $21\, ^oC$ temperature,the volume of $212\, g$ of $O_2$ gas is $34\, dm^3$. If the pressure of the gas becomes $1.24\, bar$,then how many grams of $O_2$ gas are left in the container? $(R = 0.083\, dm^3\, bar\, K^{-1}\, mol^{-1})$

(D) Step $1$: Calculate the initial pressure $(P_1)$ using the Ideal Gas Equation $PV = nRT$.
$n_1 = \frac{212\, g}{32\, g/mol} = 6.625\, mol$.
$T = 21 + 273.15 = 294.15\, K$.
$P_1 = \frac{n_1RT}{V} = \frac{6.625 \times 0.083 \times 294.15}{34} = 4.76\, bar$.
Step $2$: Calculate the new amount of gas $(n_2)$ at $P_2 = 1.24\, bar$ using $PV = nRT$ (assuming $V$ and $T$ are constant).
$n_2 = \frac{P_2V}{RT} = \frac{1.24 \times 34}{0.083 \times 294.15} = 1.727\, mol$.
Step $3$: Convert moles to mass.
$Mass = n_2 \times Molar\, mass = 1.727 \times 32 = 55.26\, g$.
(Note: The provided solution $156.74\, g$ appears to be incorrect based on the calculation; the correct remaining mass is $55.26\, g$).