[" A water sample contains "9.5%MgCl_(2)" and "11.7%],[" NaCl (by weight).Assuming "80%" ionisation of each "],[" salt.Bolling point of water will be approximately "],[(K_(b)=0.52)]

A water sample contains 9.5% MgCl_(2) and 11.7% NaCl (by weight).Assuming 80% ionisation of each salt boiling point of water will be (K_(b)=0.52)

A water sample contains 9.5% MgCl_(2) and 11.7% NaCl (by weight).Assuming 80% ionisation of each salt boiling point of water will be (K_(b)=0.52)

A water sample contains 9.5% MgCl_2 and 11.7% NaCl (by weight). Assuming 80% ionisation of each salt boiling point of water will be ( K_b = 0.52 )

K_(b) for water is 0.52 K/m. Then, 0.1 m solution of NaCl will boil approximately at

River water is found to contain 11.7% NaCl,9.5% "MgCl"_(2), and 8.4%. "NaHCO"_(3) , by weight of solution. Calculate its normal boiling point assuming 90% ionization of NaCl, 70% ionization of "MgCl"_(2) and 50% ionization of "NaHCO"_(3)("K"_("b") " for water" =0.52)

Sea water is found to contain 5.85% NaCI and 9.50% MgCI_(2) by weight of solution. Calculate its normal boiling point assuming 80% ionisation for NaCI and 50% ionisation of MgCI[K_(b)(H_(2)O) =0.51 kg mol^(-1)K]

Sea water is found to contain 5.85% NaCI and 9.50% MgCI_(2) by weight of solution. Calculate its normal boiling point assuming 80% ionisation for NaCI and 50% ionisation of MgCI[K_(b)(H_(2)O) =0.51 kg mol^(-1)K]

Sea water is found to contain 5.85% NaCI and 9.50% MgCI_(2) by weight of solution. Calculate its normal boiling point assuming 80% ionisation for NaCI and 50% ionisation of MgCI[K_(b)(H_(2)O) =0.51 kg mol^(-1)K]

Each cell contains 50% water by weight.