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Q.
If 15 g of a solute in 100 g of water makes the solution freeze at $ -1{}^\circ C $ , then 30 g. of the same solute in 100 g of water will make the solution freeze at:
We know that, $ m=\frac{1000{{K}_{f}}.w}{\Delta T\cdot W} $ (As the solute and solvent are same in both cases), hence: $ \frac{{{W}_{1}}}{{{W}_{1}}}=\Delta {{T}_{1}} $ and $ \frac{{{W}_{2}}}{{{W}_{2}}}=\Delta {{T}_{2}} $ or $ \frac{{{W}_{1}}\cdot {{W}_{2}}}{{{W}_{1}}\cdot {{W}_{2}}}=\frac{\Delta {{T}_{1}}}{\Delta {{T}_{2}}} $ $ \frac{15\times 100}{100\times 30}=\frac{[(0)-(-1)]}{\Delta {{T}_{2}}} $ $ \Delta {{T}_{2}}=2\times 1=2{}^\circ $ Hence, now water will freeze at $ (0-2)=-2{}^\circ C $