A) \[-\,2{}^\circ C\]
B) \[-\,0.5{}^\circ C\]
C) \[0{}^\circ C\]
D) \[2{}^\circ C\]
Correct Answer: A
Solution :
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\[(0-2)=-2{}^\circ C\]
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