Question:
A solution containing \(6.0\;g\) of urea is isotonic with a solution containing \(10\;g\) of a non-electrolytic solute \(X\). The molar mass of \(X\) (in \(g\;mol^{-1}\)) is
A solution containing \(6.0\;g\) of urea is isotonic with a solution containing \(10\;g\) of a non-electrolytic solute \(X\). The molar mass of \(X\) (in \(g\;mol^{-1}\)) is
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For isotonic non-electrolyte solutions at the same temperature, the molar concentrations are equal. First calculate moles of one solute, then use the same moles for the second solution.
Updated On: Jun 22, 2026
- \(50.0\)
- \(100\)
- \(75.0\)
- \(68.0\)
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The Correct Option is B
Solution and Explanation
Step 1: Understand the meaning of isotonic solutions.
Two solutions are said to be isotonic when they have the same osmotic pressure.
For dilute solutions, osmotic pressure is given by:
\[ \pi = CRT \] where,
\[ C = \text{molar concentration} \] \[ R = \text{gas constant} \] \[ T = \text{absolute temperature} \] Since both solutions are isotonic and measured at the same temperature, their osmotic pressures are equal.
Therefore:
\[ C_1=C_2 \] This means the number of moles of solute present in equal volumes of solution must be equal.
Step 2: Calculate the number of moles of urea.
Urea has molecular formula:
\[ NH_2CONH_2 \] Molar mass of urea is:
\[ 12 + 16 + 2(14) + 4(1) \] \[ =60\;g\;mol^{-1} \] Given mass of urea:
\[ 6.0\;g \] Moles of urea are:
\[ \text{Moles}=\frac{\text{Mass}}{\text{Molar mass}} \] \[ =\frac{6.0}{60} \] \[ =0.1\;mol \]
Step 3: Apply isotonic condition.
Since both solutions are isotonic and both solutes are non-electrolytes, the number of particles in solution must be equal.
Hence, moles of solute \(X\) are also:
\[ 0.1\;mol \]
Step 4: Calculate molar mass of solute \(X\).
Given mass of solute \(X\):
\[ 10\;g \] Using the formula:
\[ \text{Molar mass}=\frac{\text{Mass}}{\text{Moles}} \] \[ =\frac{10}{0.1} \] \[ =100\;g\;mol^{-1} \]
Step 5: Match with the given options.
The correct option is:
\[ (2)\;100 \]
Step 6: Final conclusion.
Hence, the molar mass of solute \(X\) is:
\[ \boxed{100\;g\;mol^{-1}} \]
Two solutions are said to be isotonic when they have the same osmotic pressure.
For dilute solutions, osmotic pressure is given by:
\[ \pi = CRT \] where,
\[ C = \text{molar concentration} \] \[ R = \text{gas constant} \] \[ T = \text{absolute temperature} \] Since both solutions are isotonic and measured at the same temperature, their osmotic pressures are equal.
Therefore:
\[ C_1=C_2 \] This means the number of moles of solute present in equal volumes of solution must be equal.
Step 2: Calculate the number of moles of urea.
Urea has molecular formula:
\[ NH_2CONH_2 \] Molar mass of urea is:
\[ 12 + 16 + 2(14) + 4(1) \] \[ =60\;g\;mol^{-1} \] Given mass of urea:
\[ 6.0\;g \] Moles of urea are:
\[ \text{Moles}=\frac{\text{Mass}}{\text{Molar mass}} \] \[ =\frac{6.0}{60} \] \[ =0.1\;mol \]
Step 3: Apply isotonic condition.
Since both solutions are isotonic and both solutes are non-electrolytes, the number of particles in solution must be equal.
Hence, moles of solute \(X\) are also:
\[ 0.1\;mol \]
Step 4: Calculate molar mass of solute \(X\).
Given mass of solute \(X\):
\[ 10\;g \] Using the formula:
\[ \text{Molar mass}=\frac{\text{Mass}}{\text{Moles}} \] \[ =\frac{10}{0.1} \] \[ =100\;g\;mol^{-1} \]
Step 5: Match with the given options.
The correct option is:
\[ (2)\;100 \]
Step 6: Final conclusion.
Hence, the molar mass of solute \(X\) is:
\[ \boxed{100\;g\;mol^{-1}} \]
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