18 text{ g} of glucose (molar mass = 180 text | vedclass

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(B) Step $1$: Calculate the number of moles of glucose $(n)$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{18 	ext{ g}}{180 	ext{ g/mol}} = 0.

Vedclass · Accepted Answer

$4.73$

Vedclass · Answer

(B) Step $1$: Calculate the number of moles of glucose $(n)$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{18 	ext{ g}}{180 	ext{ g/mol}} = 0.1 	ext{ mol}$.Step $2$: Convert volume to liters $(V)$: $V = 500 	ext{ ml} = 0.5 	ext{ L}$.Step $3$: Convert temperature to Kelvin $(T)$: $T = 15 + 273.15 = 288.15 	ext{ K}$.Step $4$: Use the osmotic pressure formula $\pi = CRT = \frac{n}{V}RT$: $\pi = \frac{0.1 	ext{ mol}}{0.5 	ext{ L}} 	imes 0.0821 	ext{ L atm K}^{-1} 	ext{ mol}^{-1} 	imes 288.15 	ext{ K} = 0.2 	imes 0.0821 	imes 288.15 = 4.732 	ext{ atm}$.Thus, the osmotic pressure is approximately $4.73 	ext{ atm}$.

$18 \text{ g}$ of glucose (molar mass $= 180 \text{ g/mol}$) is dissolved in water to prepare $500 \text{ ml}$ solution at $15^{\circ}\text{C}$. Calculate the osmotic pressure of the solution. $[R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}]$ (in $\text{ atm}$)

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