Molar mass of ethylene glycol (C$_2$H$_6$O$_2$) $= 62\ \text{g mol}^{-1}$

Moles of glycol $= \dfrac{62}{62} = 1\ \text{mol}$

Let $w$ g of water freeze. Remaining water = $(250 - w)\ \text{g}$.

At $-10°$C: $\Delta T_f = 10$

$m = \dfrac{1}{(250-w)/1000} = \dfrac{1000}{250-w}$

$10 = 1.86 \times \dfrac{1000}{250-w}$

$250 - w = \dfrac{1860}{10} = 186$

$w = 250 - 186 = \mathbf{64\ \text{g}}$

Wait — that gives 64 g (option C, index 2). Let me recheck: $250-w=186$, $w=64$. Official answer is 48 g. Using $w=48$: $250-48=202$, molality $=1000/202=4.95$, $\Delta T_f=1.86\times4.95=9.2\neq10$. The correct answer is indeed 64 g (index 2).

$LC = \frac{0.5}{100} = 0.005 \text{ mm}$.

Zero Error: Since the zero is 3 divisions below the mean line, it is a positive zero error. Zero Error $= +3 \times 0.005 = +0.015 \text{ mm}$.

Observed Reading: $5.5 + (48 \times 0.005) = 5.5 + 0.240 = 5.740 \text{ mm}$.

Actual Thickness: Observed Reading $-$ Zero Error $= 5.740 - 0.015 = 5.725 \text{ mm}$.

Rewrite: $\frac{d}{dx}(y\sin x) = 4x$

Integrate: $y\sin x = 2x^2 + C$

At $x = \frac{\pi}{2}, y=0$: $0 = 2\cdot\frac{\pi^2}{4} + C \Rightarrow C = -\frac{\pi^2}{2}$

So $y = \frac{2x^2 - \frac{\pi^2}{2}}{\sin x}$

At $x = \frac{\pi}{6}$: $\sin\frac{\pi}{6} = \frac{1}{2}$, so $y = \frac{2\cdot\frac{\pi^2}{36} - \frac{\pi^2}{2}}{\frac{1}{2}} = 2\left(\frac{\pi^2}{18} - \frac{\pi^2}{2}\right) = 2\cdot\frac{\pi^2 - 9\pi^2}{18} = \frac{-8\pi^2}{9} \cdot \frac{1}{2\sin(\pi/6)}$

Final answer: $y\!\left(\frac{\pi}{6}\right) = \mathbf{-\dfrac{8\pi^2}{9\sqrt{3}}}$