Question:easy

At \(27^{\circ}C\), x g of \(C_6H_{12}O_6\) (molar mass = \(180 \text{ g mol}^{-1}\)) and y g of a non-volatile, non-electrolyte (molar mass = \(92 \text{ g mol}^{-1}\)) were present separately in 1.0 L solutions. The osmotic pressure of two solutions is equal. What is x/y?

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When osmotic pressures are equal at the same temperature, the molarity of the solute particles must be identical.
Updated On: Jun 9, 2026
  • \(45/23\)
  • \(23/45\)
  • \(32/54\)
  • \(54/32\)
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The Correct Option is A

Solution and Explanation

Step 1: Identify the property that is equal.
Both solutions have the same osmotic pressure at the same temperature $27^{\circ}C$. Osmotic pressure is a colligative property given by $\Pi = CRT$, so if $\Pi$ and $T$ are equal, the molar concentrations $C$ must be equal too.
Step 2: Note that both volumes are 1.0 L.
Since both solutions are made in $1.0$ L, molarity is simply moles divided by $1$, which equals the number of moles. So equal concentration means equal moles of solute.
Step 3: Write moles in terms of given masses.
Moles of glucose $= \dfrac{x}{180}$ and moles of the other solute $= \dfrac{y}{92}$.
Step 4: Set the moles equal.
Because the concentrations match, \[ \frac{x}{180} = \frac{y}{92} \]
Step 5: Rearrange to get the ratio x/y.
Cross multiplying gives \[ \frac{x}{y} = \frac{180}{92} \]
Step 6: Simplify the fraction.
Dividing top and bottom by $4$, we get $\dfrac{180}{92} = \dfrac{45}{23}$. So the required ratio is $45/23$.
\[ \boxed{\dfrac{45}{23}} \]
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