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A $1.17 \%$ solution of solute $A$ is isotonic with $7.2 \%$ solution of glucose. If the molecular weight of solute $A$ is 58.5, the value of van't Hoff factor, ' $i$ ' is
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2
$\because$ Molar mass of glucose $(M)=180 \mathrm{~g}$
Molar mass of solute $(A)$, i.e. $\left(M_A\right)=58.5$
$\%$ of $A=1.17 \%$
$\%$ of glucose $=7.2 \%$
Also,
$\because \pi(\text { osmotic pressure })=C R T$
$\pi=\frac{n}{V} \cdot R T \Rightarrow \pi=\frac{w}{M} \times R T$
where, $\pi=$ osmotic pressure, $w=\%$ of sample
$M=$ molar mass,$R=$ gas constant
$T=$ temperature
$\because$ These are isotonic.
$\therefore \quad i=\frac{M_A \times \% \text { of glucose }}{\% \text { of } A \times M}$
$i=\frac{58.5 \times 7.2}{1.17 \times 180}=\frac{421.2}{210.6}=2$
Molar mass of solute $(A)$, i.e. $\left(M_A\right)=58.5$
$\%$ of $A=1.17 \%$
$\%$ of glucose $=7.2 \%$
Also,
$\because \pi(\text { osmotic pressure })=C R T$
$\pi=\frac{n}{V} \cdot R T \Rightarrow \pi=\frac{w}{M} \times R T$
where, $\pi=$ osmotic pressure, $w=\%$ of sample
$M=$ molar mass,$R=$ gas constant
$T=$ temperature
$\because$ These are isotonic.
$\therefore \quad i=\frac{M_A \times \% \text { of glucose }}{\% \text { of } A \times M}$
$i=\frac{58.5 \times 7.2}{1.17 \times 180}=\frac{421.2}{210.6}=2$
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