Sea water contains 29.25% NaCl and 19% MgCl_2 | vedclass

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(D) Mass of solution = $100 \, g$.Mass of $NaCl = 29.25 \, g$,Mass of $MgCl_2 = 19 \, g$.Mass of solvent $(H_2O)$ = $100 - (29.25 + 19) = 51.75 \, g =

Vedclass · Accepted Answer

$116$

Vedclass · Answer

(D) Mass of solution = $100 \, g$.Mass of $NaCl = 29.25 \, g$,Mass of $MgCl_2 = 19 \, g$.Mass of solvent $(H_2O)$ = $100 - (29.25 + 19) = 51.75 \, g = 0.05175 \, kg$.For $NaCl$ $(i=2)$: Moles = $29.25 / 58.5 = 0.5 \, mol$.For $MgCl_2$ $(i=3)$: Moles = $19 / 95 = 0.2 \, mol$.Total moles of ions = $(i_{NaCl} 	imes n_{NaCl}) + (i_{MgCl_2} 	imes n_{MgCl_2}) = (2 	imes 0.5) + (3 	imes 0.2) = 1.0 + 0.6 = 1.6 \, mol$.$\Delta T_b = i 	imes K_b 	imes m = K_b 	imes (n_{total_ions} / 	ext{mass of solvent in kg}) = 0.52 	imes (1.6 / 0.05175) \approx 16.07 \, ^{\circ}C$.Boiling point of solution = $100 + 16.07 = 116.07 \, ^{\circ}C$.Nearest integer is $116$.

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