The amount of wet NaOH containing 15% water required to prepare 70 liters of 0.5 N solution is:

Options

A. (a)1.65 kg
B. (b)1.4 kg
C. (c )16.5 kg
D. (d)140 kg

Answer

A

Explanation

Moles of pure NaOH required`=(NxxV_(L))/(v.f)=(0.5xx70)/1=35`
Mass of pure `NaOH` required`=35xx40=1400 g=1.4 kg`.
Mass of wet `NaOH` required (containing 15 % water)
`=1.4xx100/85=1.65 kg`.