The correct option is A 2
Let us assume the weight of acetic acid present in solution be w g.
NaOH+CH3COOH→CH3COONa+H2O
Mole fraction of acetic acid in solution = xacetic acid=0.18
xacetic acid = nacetic acidnacetic acid+nwater
xacetic acid = w/60w/60+(71−w)/18 = 0.18
On solving, w = 30 g
moles of acetic acid = given massmolar mass = 3060 = 0.5 mol
For complete neutralization,
Number of moles of NaOH = number of moles of CH3COOH
1×V=0.5
Where, V is voume of NaOH required
Since, V=0.5/1=1/x
∴x=2