2 mole each of $C{H}_{3}OH$ & $C{H}_{3}COOH$ are taken and heated in the presence of conc. $v$ so that equilibrium is established. If $K_c$ for esterification process is 4, mass of ester present at equilibrium (in gm) is

 Reaction is:   $C{H}_{3}OH+C{H}_{3}COOH\rightleftharpoons C{H}_{3}COOC{H}_3$ in $H_{2}S{O}_4$ presence

Initial moles:  2 mol + 2 mol                         

Moles at equilibrium: $(2-x)mol+(2-x)mol\rightleftharpoons xmol+xmol$

Conc. at equilibrium: $(2-x)/V+(2-x)/V\rightleftharpoons x/V+ x/V$

$K_c=\left[C{H}_{3}COOC{H}_3\right]\left[H_{2}O\right]/\left[C{H}_{3}OH\right]\left[C{H}_{3}COOH\right]=(x/V)(x/V)/\left[\right(2-x)/V]\left[\right(2-x)/V]=4$

$x^2/(2-x{)}^2=4;x^2=4(2-x{)}_2;x=2(2-x);3x=4;x=4/3$

therefore, no. of moles of CH3COOCH3 at equilibrium= 4/3 , since molar mass of ester is 74 g/mol1mole of ester = 74g, then 4/3 moles of ester contains = (74)4/3 =98.6g ≈98.4g                                                 

Hence, the correct option A is i.e., 98.4