Question

weight of salt in ( 1 mathrm{Ky} 20 % mathrm{sin}=200 mathrm{g} )
but in table salt only ( 85 % ) is salt
thus ( 85 % ) of ( x=200 mathrm{g} )
weight of tave salt
[
begin{array}{l}
x=frac{200 x+00}{81}
x=frac{4000}{17}=235 cdot 29 y
end{array}
]

# 2. How many gram of table salt containing 15% moisture is needed to prepare 1 kg of a 20% salt solution? (A) 200 g (B) 230 g ES (C) 235 g (D) 170g Two Litre of 1.5 M-NaOH solution is mixed with three Litre of 2 M-NaOH solution and one Litre water is also

Solution