(B) The osmotic pressure formula is given by $\pi = \frac{n}{V} RT$. For the initial state: $\pi = x = \frac{1}{10.5} RT$. For the final state: $\pi'

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(B) The osmotic pressure formula is given by $\pi = \frac{n}{V} RT$. For the initial state: $\pi = x = \frac{1}{10.5} RT$. For the final state: $\pi' = \frac{x}{10} = \frac{1}{V'} RT$. Dividing the two equations: $\frac{x}{x/10} = \frac{\frac{1}{10.5} RT}{\frac{1}{V'} RT}$. This simplifies to $10 = \frac{V'}{10.5}$. Therefore,$V' = 10 \times 10.5 = 105 \ m^{3}$.

Class 12 Chemistry Solutions Osmosis and Osmotic pressure of the solution

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$A$ solution has an osmotic pressure of $x \ kPa$ at $300 \ K$ having $1 \ mole$ of solute in $10.5 \ m^{3}$ of solution. If its osmotic pressure is reduced to $(\frac{1}{10})$th of its initial value,what is the new volume of solution (in $m^{3}$)?

MHT CET 2020 All MHT CET

A
$30$
B
$105$
C
$110$
D
$11.0$

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$A$ solution has an osmotic pressure of $x \ kPa$ at $300 \ K$ having $1 \ mole$ of solute in $10.5 \ m^{3}$ of solution. If its osmotic pressure is reduced to $(\frac{1}{10})$th of its initial value,what is the new volume of solution (in $m^{3}$)?

Similar Questions

The osmotic pressure of a $1 \ M$ solution at $27^{\circ}C$ is $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$: (in $atm$)

$A$ solution containing $500 \ g$ of protein per liter of solution is isotonic with a solution containing $3.42 \ g$ of sugar per liter of solution. The molar mass of the protein is:

$A$ solution containing $10 \ g$ per $dm^3$ of urea (molecular mass $= 60 \ g \ mol^{-1}$) is isotonic with a $5 \%$ solution of a non-volatile solute. The molecular mass of this non-volatile solute is ........ $g \ mol^{-1}$.

Which one has the highest osmotic pressure?

The relationship between osmotic pressure at $273 \ K$ when $10 \ g$ glucose $(P_1)$,$10 \ g$ urea $(P_2)$,and $10 \ g$ sucrose $(P_3)$ are dissolved in $250 \ mL$ of water is:

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