(A) The formula for osmotic pressure is $\pi = CRT$. First,calculate the number of moles of cane sugar: $n = \frac{\text{mass}}{\text{molar mass}} = \

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(A) The formula for osmotic pressure is $\pi = CRT$. First,calculate the number of moles of cane sugar: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{34.2 \ g}{342 \ g \ mol^{-1}} = 0.1 \ mol$. The concentration $C = \frac{n}{V} = \frac{0.1 \ mol}{1 \ L} = 0.1 \ mol \ L^{-1}$. The temperature $T = 20 + 273 = 293 \ K$. Substituting the values: $\pi = 0.1 \ mol \ L^{-1} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 293 \ K = 2.4026 \ atm \approx 2.40 \ atm$.

Class 12 Chemistry Solutions Osmosis and Osmotic pressure of the solution

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The osmotic pressure of a solution containing $34.2 \ g$ of cane sugar (molar mass = $342 \ g \ mol^{-1}$) in $1 \ L$ of solution at $20^{\circ} C$ is
(Given,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$) (in $atm$)

MHT CET 2017 All MHT CET

A
$2.40$
B
$3.6$
C
$24$
D
$0.0024$

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Osmosis and Osmotic pressure of the solution

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The osmotic pressure of a solution containing $34.2 \ g$ of cane sugar (molar mass = $342 \ g \ mol^{-1}$) in $1 \ L$ of solution at $20^{\circ} C$ is (Given,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$) (in $atm$)

Similar Questions

Which of the following pairs of solutions can be isotonic at the same temperature?

$A$ $5\% \ (w/v)$ solution of anhydrous $CaCl_2$ at $0^o C$ exerts an osmotic pressure of $15 \ atm$. The degree of dissociation $(\alpha)$ of $CaCl_2$ is ....... $\%$.

To have an osmotic pressure equal to a solution containing $9.2 \ g$ of glucose per liter,how many grams of glucose must be dissolved in $0.5 \ L$ of solution?

In which case does osmosis not occur from solution $A$ to solution $B$?

Calculate the osmotic pressure exerted by a solution containing $0.822 \ g$ of solute in $300 \ mL$ of water at $300 \ K$. (Molar mass of solute $= 340 \ g \ mol^{-1}, R = 0.0821 \ L \ atm \ mol^{-1} \ K^{-1}$) (in $atm$)

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The osmotic pressure of a solution containing $34.2 \ g$ of cane sugar (molar mass = $342 \ g \ mol^{-1}$) in $1 \ L$ of solution at $20^{\circ} C$ is
(Given,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$) (in $atm$)