A solution of 17.1 w% of sucrose (molar mass | vedclass

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Question

(C) Given,Mass of sucrose $(w_1) = 17.1 \ g$ Molar mass of sucrose $(M_1) = 342 \ g \ mol^{-1}$ Mass of oxalic acid $(w_2) = x \ g$ Molar mass of oxal

Vedclass · Accepted Answer

$4.41$

Vedclass · Answer

(C) Given,Mass of sucrose $(w_1) = 17.1 \ g$ Molar mass of sucrose $(M_1) = 342 \ g \ mol^{-1}$ Mass of oxalic acid $(w_2) = x \ g$ Molar mass of oxalic acid $(M_2) = 90 \ g \ mol^{-1}$ Degree of dissociation $(\alpha)$ of oxalic acid $= 0.01$ Step $I$: To find the value of van't Hoff factor $(i)$ $i = (1 - \alpha) + n\alpha$ where, $n = 3$ (number of ions produced by oxalic acid, $H_2C_2O_4 ightleftharpoons 2H^+ + C_2O_4^{2-}$) $i = (1 - 0.01) + (3 	imes 0.01) = 0.99 + 0.03 = 1.02$ Step $II$: For isotonic solutions, osmotic pressure $(\pi)$ is equal. $\pi_{	ext{sucrose}} = \pi_{	ext{oxalic acid}}$ $C_{	ext{sucrose}} 	imes R 	imes T = i 	imes C_{	ext{oxalic acid}} 	imes R 	imes T$ $\frac{w_1}{M_1} = i 	imes \frac{x}{M_2}$ $x = \frac{w_1 	imes M_2}{M_1 	imes i} = \frac{17.1 	imes 90}{342 	imes 1.02} = \frac{1539}{348.84} \approx 4.41$ Thus, $x = 4.41$.

$A$ solution of $17.1$ $w\%$ of sucrose (molar mass $= 342 \ g \ mol^{-1}$) is isotonic with a $x$ $w\%$ solution of oxalic acid (molar mass $= 90 \ g \ mol^{-1}$). Assume the degree of dissociation of oxalic acid as $0.01$. What is $x$?

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