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Daniel [21]
3 years ago
6

An aqueous solution containing 17.5 g of an unknown molecular (nonelectrolyte) compound in 100.0 g of water has a freezing point

of -1.8 degrees C. Calculate the molar mass of the unknown compound.
Chemistry
1 answer:
Sonbull [250]3 years ago
6 0

Answer:

molar mass = 180.833 g/mol

Explanation:

  • mass sln = mass solute + mass solvent

∴  solute: unknown molecular (nonelectrolyte)

∴ solvent: water

∴ mass solute = 17.5 g

∴ mass solvent =  100.0 g = 0.1 Kg

⇒ mass sln = 117.5 g

freezing point:

  • ΔTc = - Kc×m

∴ ΔTc = -1.8 °C

∴ Kc H2O = 1.86 °C.Kg/mol

∴ m: molality (mol solute/Kg solvent)

⇒ m = ( - 1.8 °C)/( - 1.86 °C.Kg/mol)

⇒ m = 0.9677 mol solute/Kg solvent

  • molar mass (Mw) [=] g/mol

∴ mol solute = ( m )×(Kg solvent)

⇒ mol solute = ( 0.9677 mol/Kg) × ( 0.100 Kg H2O )

⇒ mol solute = 0.09677 mol

⇒ Mw solute = ( 17.5 g ) / ( 0.09677 mol )

⇒ Mw solute = 180.833 g/mol

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M_{I^-}=0.6841M

Explanation:

Hello.

In this case, since the precipitation of the lead iodide is related to the iodide ion in solution, if we make react lead (II) nitrate with an iodide-containing salt, a possible chemical reaction would be:

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n_{I^-}=0.01393molPb(NO_3)_2*\frac{2molI^-}{1molPb(NO_3)_2} =0.02785molI^-

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