Question

Equal volume of 0.2 N Na2SO4 and 0.1 N BaCl2 solutions are mixed together. Assume that BaSO4 is completely insoluble. If Kb(H2O) = 0.52 K kg mol^–1, what would be the normal boiling point of the resulting solution? (Assume molality = molarity) a)100.15°C b)100.75°C c)100.091°C d)100.175°C

Answer 1

Answer:

Explanation:

100°C

132k

ok this won't chang thank you guys

Answer 2

Given,

0.2N Na$_2$SO$_4$ and 0.1N BaCl$_2$ mixed in equal volumes

Kb(H$_2$O) = 0.52 K kg $mol^{-1}$

BaSO$_4$ is completely soluble

To find,

The boiling point of the resulting solution

Solution,

During the reaction,

Na$_2$SO$_4$    +     BaCl$_2$    ---------->    BaSO$_4$    +    2NaCl

  0.2N              0.1N

  0.1M               0.05M

0.1×V               0.05×V                    0                      0      mol

0.05V               0                           0.05V              2×0.05     mol

Total mol of ion = 0.05V×3 + 2×0.05V×2

                          = 0.15V + 0.2V

                          = 0.35V

molarity = mol/V = 0.35V/2V

                           = 0.175 V

change in boiling point,

ΔT$_b$ = K$_b$ m

      = 0.52 × 0.175

      = 0.091°C

∴ Normal boiling point = 100°C +0.091°C = 100.091°C (option c)