Answer to Question #162690 in General Chemistry for Sashi

Question

Calculate the molality of 9.46 g ethylene glycol (C₂H₆O₂) in 360 g water. 

Answer 1

b_{(ethylene glycol)} = n_{(ethylene glycol)} / m_{(H2O, kg)}

n_{(ethylene glycol)} = m_{(ethylene glycol)} / M_{(ethylene glycol)} = 9.46 / 62 = 0.153 (mol)

b_{(ethylene glycol)} = 0.153 / 0.36 = 0.425 (mol/kg)

Answer: b = 0.425 (mol/kg).