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Answer 1 · 2014-12-15T12:30:49

The answer is ${0.56}^{\circ} C$ $0.56^@C$ .

Boiling-point elevation is modeled by this equation

$DeltaT_b = k_b * m$ , where

$DeltaT_b$ - represents the boiling-point elevation;
$k_b$ - represents water's ebullioscopic constant;
$m$ - represents the solution's molality;

SInce we are dealing with water, which has a density of $rho = 1.0(kg)/L$ , the mass of the solvent will be

$m_(water) = rho_(water) * V = 1.0 (kg)/L * 2.0L = 2.0 kg$

So, the solution's molality wil be

$m = (2.1 mol es)/(2.0 kg) = 1.1m$

Therefore, the boiling-point elevation will be

$DeltaT_b = 0.51^@C/m * 1.1m = 0.56^@C$