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3 years ago

What is the change in boiling point for a 0.615m solution of Mgl2 in water?

Chemistry

1 answer:

ivanzaharov [21]3 years ago

4 0

Answer : The change in boiling point is, $0.94^oC$

Explanation :

Formula used :

$\Delta T_b=i\times K_f\times m$

where,

$\Delta T_b$ = change in boiling point = ?

i = Van't Hoff factor = 3 (for MgI₂ electrolyte)

$K_f$ = boiling point constant for water = $0.51^oC/m$

m = molality = 0.615 m

Now put all the given values in this formula, we get

$\Delta T_b=3\times (0.51^oC/m)\times 0.615m$

$\Delta T_b=0.94^oC$

Therefore, the change in boiling point is, $0.94^oC$

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<h3>Further explanation</h3>

Given

1.72 moles of NOCI

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