when-9-5-g-of-lithium-stearate-lic18h35o2-mw-290-47-g-mol-is-dissolved-in-550-g-

Question

When 9.5 g of lithium stearate (LiC₁₈H₃₅O₂, MW = 290.47 g/mol) is dissolved in 550 g benzene, the resulting solution freezes at 0.30 °C lower than pure benzene (FP = 5.5 °C). Calculate the value of Kf (freezing point constant) for benzene.

Accepted Answer

K_f = 5.0 °C•kg/mol

Answer

K_f = 2.8 °C•kg/mol

Answer

K_f = 4.6 °C•kg/mol

Answer

K_f = 5.4 °C•kg/mol

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When 9.5 g of lithium stearate (LiC₁₈H₃₅O₂, MW = 290.47 g/mol) is dissolved in 550 g benzene, the resulting solution freezes at 0.30 °C lower than pure benzene (FP = 5.5 °C). Calculate the value of Kf (freezing point constant) for benzene.

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When 9.5 g of lithium stearate (LiC18H35O2, MW = 290.47 g/mol) is dissolved in 550 g benzene, the resulting solution freezes at 0.30 °C lower than pure benzene (FP = 5.5 °C). Calculate the value of Kf (freezing point constant) for benzene.

When 9.5 g of lithium stearate (LiC₁₈H₃₅O₂, MW = 290.47 g/mol) is dissolved in 550 g benzene, the resulting solution freezes at 0.30 °C lower than pure benzene (FP = 5.5 °C). Calculate the value of Kf (freezing point constant) for benzene.