A 1.17 % solution of solute A is isotonic wit | vedclass

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(A) For isotonic solutions, the osmotic pressure $\pi$ is equal, so $\pi_1 = \pi_2$. Since $\pi = i \cdot C \cdot R \cdot T$, and $C = \frac{w \% 	im

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$2$

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(A) For isotonic solutions, the osmotic pressure $\pi$ is equal, so $\pi_1 = \pi_2$. Since $\pi = i \cdot C \cdot R \cdot T$, and $C = \frac{w \% 	imes 10}{M}$, we have $i_1 \cdot \frac{w_1 \%}{M_1} = i_2 \cdot \frac{w_2 \%}{M_2}$. For glucose, the van't Hoff factor $i_2 = 1$ and molar mass $M_2 = 180 \ g/mol$. Given $w_1 \% = 1.17$, $M_1 = 58.5$, and $w_2 \% = 7.2$. Substituting the values: $i_1 	imes \frac{1.17}{58.5} = 1 	imes \frac{7.2}{180}$. $i_1 	imes 0.02 = 0.04$. $i_1 = \frac{0.04}{0.02} = 2$.

$A$ $1.17 \%$ solution of solute $A$ is isotonic with a $7.2 \%$ solution of glucose. If the molecular weight of solute $A$ is $58.5$, the value of the van't Hoff factor,'$i$',is:

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