IF 10 g of solute was dissolved in 250 mL. of water and osmotic pressure of the solution was found to be 600 mm of Hg at 300 K, then molecular weight of the solute is _______.`g mol ^(-1)`

Options

A. `(22800xxR)/(1.5)`
B. `(22800)/(1.5)`
C. `(1.5xxR)/(22800)`
D. `(228xxR)/(1.5)`

Answer

To find the molecular weight of the solute using the given data, we can use the formula for osmotic pressure:

\[ \pi = CRT \]

Where:
- \(\pi\) = osmotic pressure (in atm)
- \(C\) = concentration of the solution (in mol/L)
- \(R\) = ideal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature (in K)

### Step 1: Convert Osmotic Pressure to atm
Given that the osmotic pressure \(\pi\) is 600 mm of Hg, we need to convert this to atm:

\[ \pi = \frac{600 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 0.7895 \text{ atm} \]

**Hint:** Remember that 1 atm = 760 mm Hg for conversion.

### Step 2: Calculate the Concentration (C)
The concentration \(C\) can be expressed in terms of moles of solute and volume of solution:

\[ C = \frac{n}{V} \]

Where:
- \(n\) = number of moles of solute
- \(V\) = volume of solution in liters

Given that 10 g of solute is dissolved in 250 mL of water, we convert 250 mL to liters:

\[ V = \frac{250 \text{ mL}}{1000} = 0.250 \text{ L} \]

**Hint:** Always convert volumes to liters when using the ideal gas law.

### Step 3: Relate Moles to Mass and Molecular Weight
The number of moles \(n\) can be expressed as:

\[ n = \frac{W}{M} \]

Where:
- \(W\) = mass of solute (10 g)
- \(M\) = molecular weight of the solute (g/mol)

Substituting this into the concentration formula gives:

\[ C = \frac{W}{MV} \]

### Step 4: Substitute into the Osmotic Pressure Equation
Substituting \(C\) into the osmotic pressure equation:

\[ \pi = \frac{W}{MV}RT \]

Rearranging for \(M\):

\[ M = \frac{WRT}{\pi V} \]

### Step 5: Substitute Known Values
Now we can substitute the known values into the equation:

- \(W = 10 \text{ g}\)
- \(R = 0.0821 \text{ L·atm/(K·mol)}\)
- \(T = 300 \text{ K}\)
- \(\pi = 0.7895 \text{ atm}\)
- \(V = 0.250 \text{ L}\)

\[ M = \frac{10 \times 0.0821 \times 300}{0.7895 \times 0.250} \]

Calculating the numerator:

\[ 10 \times 0.0821 \times 300 = 246.3 \]

Calculating the denominator:

\[ 0.7895 \times 0.250 = 0.197375 \]

Now, substituting these values back into the equation for \(M\):

\[ M = \frac{246.3}{0.197375} \approx 1247.3 \text{ g/mol} \]

### Conclusion
The molecular weight of the solute is approximately **1247.3 g/mol**.

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