Answer
$Molality = \frac{Moles of solute}{Kg\ of\ solvent}$
And, $Moles = \frac{mass}{molar mass}$
a) Sucrose moles = $\frac{14.3g}{342.3g/mol} \approx 0.0418 mol$
Sucrose molality = $\frac{0.0418mol}{676 g of water} * \frac{1000g}{1kg} \approx0.0618M$
b) Ethylene glycol molality = $\frac{7.20mol}{3546g water} * \frac{1000g}{1kg} \approx2.03M$
Work Step by Step
$Molality = \frac{Moles of solute}{Kg\ of\ solvent}$
And, $Moles = \frac{mass}{molar mass}$
a) Sucrose moles = $\frac{14.3g}{342.3g/mol} \approx 0.0418 mol$
Sucrose molality = $\frac{0.0418mol}{676 g of water} * \frac{1000g}{1kg} \approx0.0618M$
b) Ethylene glycol molality = $\frac{7.20mol}{3546g water} * \frac{1000g}{1kg} \approx2.03M$
