Calculate the amount of CaCl_2 (van't Hoff factor, i = 2.47) diss… - Vidyadip

Question

Calculate the amount of $CaCl_2 ($van't Hoff factor, $i = 2.47)$ dissolved in $2.5 L$ solution so that its osmotic pressure at $300K$ is $0.75$ atmosphere.
$[$Given: Molar mass of $CaCl_2$ is $111 \ g \ mol^{-1} , R= 0.082 \ L \ atm \ K^{-1} \ mol^{-1}.]$

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Answer

Data : $V = 2.5 L$
$T = 300 K$
$\pi=0.75 \ atm$
Molar mass of $CaCl_2 = M_2 = 111 \ g/mol$
$R = 0.082 \ L \ atm \ k^{-1} \ mol^{-1}$
Solution :
$i=\frac{\pi_{o b s}}{\pi_{c a l}}$
$2.47=\frac{0.75}{\pi_{c a l}}$
$\pi_{c a l}=0.303 \ atm$
$\pi V=n R T$
$\therefore \pi V=\frac{W_2}{M_2} \times R \times T$
$\therefore \pi_{c a l} \times V=\frac{W_2}{M_2} \times R \times T$
$\therefore W_2=\frac{\pi_{c a l} \times V \times M_2}{R \times T}$
$=\frac{0.303 \times 2.5 \times 111}{0.082 \times 300}$
$W_2=3.417 \ gm$
Mass of $CaCl_2=3.417 \ gm$

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