Colligative properties, Osmotic pressure(OP)
Question

# At 10°C, the osmotic pressure of urea solution was formed to be 500 mm. The solution is diluted 'x' times and the temperature raised to 25°C when the osmotic prssure was noticed to be 105.3mm, then 'x' is

Easy
Solution

## I(before dilution)                           II(after dilution)${\mathrm{\pi }}_{1}$= 500mm                                    ${\mathrm{\pi }}_{2}$= 105.3mmT1 = (10+273)K                                T2 = (25+273)KV1 = 1 lit                                            V2 = x lit$\large \pi =CST\\\\\frac{\pi_1}{\pi_2}=\frac{C_1T_1}{C_2T_2}$$\large \because$the amount of urea remains constant , Wt. and GMW of urea remain constant$\large \therefore \frac{\pi_1}{\pi_2} =\frac{(\frac{w}{GMW}X\frac{1}{1})XT_1}{(\frac{w}{GMW}X\frac{1}{x})XT_2}$$\large \frac{\pi_1}{\pi_2}=\frac{xT_1}{T_2}\\\frac{500}{105.3}=\frac{x(283)}{298}$$\large x=\frac{500X298}{105.3X283}$x=5$\large \therefore$the solution  is diluted by 5 times

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