Question

An ideal solution was prepared by dissolving some amount of cane sugar (non-volatile) in $0.9$ moles of water. The solution was then cooled just below its freezing temperature ($271K$), where some ice get separated out. The remaining aqueous solution registered a vapour pressure of $700$ torr at $373K$. Calculate the mass of ice separated out if the molar heat of fusion of water is $6$ kJ?

• A. $4.5g/mol$
• B. $9g/mol$
• C. $18g/mol$
• D. $36g/mol$

Answer 1

The correct option is C $18g/mol$

The molar heat of fusion of water is 6 kJ/mol.
The heat of fusion of water is 6 kJ/mol.
Hence, the number of moles of ice formed $=\frac{6kJ/mol}{6kJ/mol=1mol}$.
The mass of ice separated out $=1mol\times 18g/mol=18g/mol$