calculae the freezing point of a solution ,contaning 18 gram glucose and 68.4 gram sucrose in 200 gram of water.freezing point of pure water is 273 kelvin and cryoscopic constant for water is 1.86 K/m
1
Expert's answer
2012-12-19T10:47:05-0500
Use the formula for freezing point depression.
Change in T = -Kf x molality of solution
In this solution, molality (b) = (18 /180)/0.2 L + (68.4/342 )/0.200 L = 0.3 mol per 0.2 L = 1.5 moles/1 kg water = 1.8 moles sucrose / kg water.
Therefore,
Change in T = - 1.86 C/m * 1.5 moles/1 kg water = -2.79 C
Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…
APPROVED BY CLIENTS
"assignmentexpert.com" is professional group of people in Math subjects! They did assignments in very high level of mathematical modelling in the best quality. Thanks a lot
Comments
Leave a comment