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  • The molal freezing point constant for water is 1.86 molality. If  42.4g of Cane sugar <img alt="\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)" src="/latex-image/?%5Cleft%28%5Cma

The molal freezing point constant for water is 1.86 molality. If  42.4g of Cane sugar \left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)  are dissolved in 100g  of water, the solution will freeze at

Option: 1

-1.84^{\circ} \mathrm{C}


Option: 2

-3.6^{\circ} \mathrm{C}


Option: 3

-2.3^{\circ} \mathrm{C}


Option: 4

-4.8^{\circ} \mathrm{C}


Answers (1)

best_answer

\mathrm{\Delta T_f =\frac{1000 \times k f \times \omega}{M W} }
\mathrm{\Delta T_f =\frac{1000 \times 1.86 \times 42.4}{100 \times 342} }
\mathrm{\Delta T_f =2.3^{\circ} \ }
\mathrm{T_f =0-2.3 }
\mathrm{ =-2.3^{\circ} \mathrm{C} }.

Posted by

Suraj Bhandari

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