Equimolal solutions of NaCl, BaCl₂ and Na₃PO₄ are prepared in water, then correct statements are

‘m’ for NaCl; BaCl₂ and Na₃PO₄ is same i (for NaCl) = 2 ; i (for BaCl₂) = 3
${\left({\mathrm{\Delta T}}_{\mathrm{f}}\right)}_1=2\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{m};{\left({\mathrm{\Delta T}}_{\mathrm{f}}\right)}_2=3\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}$
Hence (1) is correct answer
For Na₃PO₄ (i = 4) and for BaCl₂ (i = 3)
∴ Freezing point is –4°C and –3°C
For BaCl2 (i = 3) and for NaCl (i = 2)
∴ Elevation in boiling point for BaCl₂ is 1.5 times higher than NaCl solution and for NaCl
(i = 2)
Na₃PO₄ (i = 4)
$\Rightarrow$Elevation in BP for NaCl is half of Na₃PO₄ solution