67200 Chemistry, Other
What is the molecular mass of a molecular compound with a freezing point -0.520 degrees C when 6.21 g is dissolved in 500 g of water?
Answer:
ΔTfr=Kmm=n/massn=m/MKwater=1.86K⋅mol−1⋅kg0.520=1.86⋅mm=0.28mol/kgn=mass⋅m=0.5⋅0.28=0.14molM=m/n=6.21/0.14=44.36g/mol
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