Question

18 g of glucose is present in 250 g of water. The freezing point of the aqueous solution of glucose is (Kf of water = 1.86 K kg mol-1)
-1.52°C
-2.5°C
-0.74°C
-1.85°C​

Answer 1

$\boxed{ \boxed{ \begin{array}{cc} \sf \: \to \: given \: : \\ \\ \hookrightarrow \sf \:mass \: of \: glucose \: \: w = 18 \: g \\ \\ \hookrightarrow \sf \:mass \: of \: water (sovent)\: , \:w_s = 250 \: g = 0.2 5\: kg \\ \\ \hookrightarrow \sf \:k_f \: of \: water = 1.86 \: k \: kg \: {mol}^{ - 1} \\ \\ \bf \: we \: know \: that, \\ \\ \red{ \hookrightarrow \sf \:molecular \: mass \: of \: glucose = M = 180} \\ \\ \bf \: and \\ \\ \pink{ \boxed{ \Delta\:T_f = k_f \times m}} \: \: \: - - - (1) \\ \\ \rm \: here \\ \hookrightarrow \sf \:\Delta\:T_f = change \: in \: freezing \: point \: \\ \\ \hookrightarrow \sf\:k_f=freezing \: point \: constant \: \\ \\ \hookrightarrow \sf \: \: m = molality \: \\ \\ \end{array}}}$

$\boxed{ \boxed{ \begin{array}{cc} \bf \: now \\ \\ \blue{ \sf \: molality \: \: m = \frac{ \sf \: number \: of \: mole(n)}{mass \: of \: solvent(w_s) \: in \: kg} } \\ \\ \sf \implies \: m = \frac{ \frac{w}{M} }{w_s} \\ \\ = \frac{ \frac{18}{180} }{0.25} \\ \\ = \frac{18}{180 \times 0.25} \\ \\ = 0.4 \: molal \\ \\ \red{\therefore \: molality \: \: m = 0.4 \: \: molal} \\ \\ now \: from \: eqn.(1) \\ \\ \\ \Delta\:T_f = k_f \times m \\ \\ = 1.86 \times 0.4 \\ \\ = 0.744 \: \sf \: k \\ \\ \bf \: \therefore \: \Delta\:T_f = 0.744 \: k \: \: \: ( \sf \: in \: kelvin \: scale) \\ \\ \\ \sf \: freezing \: point \: of \: solution = 273 - 0.744 \\ = 272.256 \: k\\ \\ \\ \sf \: convert \: it \:i nto \: \: celcius \: scale = 272.256 - 273 \\ \\ = - 0.744 {}^{ \circ} \sf\: c \: \\ \\ \sf \: (answer)\end{array}}}$

So correct answer is option C) -0.74°C