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Addition of 0.643 g of a compound to 50 ml, if benzene (density : 0.879 g/ml.) lowers the freezingpoint from 5.51°C to 5.03°C. If Kf for benzene is 5.12 K kg mol-1, calculate the molecular weight of the compound.

Shane Macguire , 10 Years ago
Grade upto college level
anser 2 Answers
Deepak Patra

Last Activity: 10 Years ago

Hello Student,
Please find the answer to your question
Give Wt. of benzene (solvent),
W = Volume * density = 50 * 0.789 = 43.95 g
Wt. of compound (solute), w = 0.643 g
Mol. wt of benzene, M = 78
Mol. wt. of solute, m = ?
Depression in freezing point, ∆Tf = 5.51 – 5.03 = 0.48
Molal freezing constant, Kf = 5.12
Now we know that,
M – 1000 *Kf *w/W *∆Tf = 1000 *5.12 *0.643/43.95 *0.48 = 156.056

Thanks
Deepak patra
askIITians Faculty

ankit singh

Last Activity: 4 Years ago

 
W = Volume * density = 50 * 0.789 = 43.95 g
Wt. of compound (solute), w = 0.643 g
Mol. wt of benzene, M = 78
Mol. wt. of solute, m = ?
Depression in freezing point, ∆Tf = 5.51 – 5.03 = 0.48
Molal freezing constant, Kf = 5.12
Now we know that,
M – 1000 *Kf *w/W *∆Tf = 1000 *5.12 *0.643/43.95 *0.48 = 156.056

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