Insulin `(C_(2)H_(10)O_(5))_(n)` is dissolved in a suitable solvent and the osmotic pressure `(pi)` of solutions of various concentrations `(g//cm^(3))C` is measured at `20^(@)C`. The slope of a plot of `pi` against `C` is found to be `4.65 xx 10^(-3)`. The molecular weight of insulin is:

Options

A. 3 x `10^(5)`
B. 9 x`10^(5)`
C. 4.5 x `10^(5)`
D. 5.16 x `10^(6)`

Answer

To find the molecular weight of insulin given the slope of the plot of osmotic pressure (π) against concentration (C), we can follow these steps:

### Step 1: Understand the relationship between osmotic pressure and concentration
The osmotic pressure (π) can be expressed using the formula:
\[ \pi = CRTI \]
where:
- \(C\) is the concentration in moles per liter (mol/L),
- \(R\) is the gas constant (0.0821 L·atm/(K·mol)),
- \(T\) is the temperature in Kelvin,
- \(I\) is the van 't Hoff factor (which is 1 for insulin as it behaves as a weak electrolyte).

### Step 2: Convert concentration from g/cm³ to mol/L
Since the concentration is given in grams per cubic centimeter (g/cm³), we need to relate it to moles. The concentration \(C\) can be expressed as:
\[ C = \frac{W}{M \cdot V} \]
where:
- \(W\) is the mass of the solute (insulin) in grams,
- \(M\) is the molecular weight in grams per mole,
- \(V\) is the volume in liters.

To convert from g/cm³ to mol/L, we note that:
1 g/cm³ = 1000 g/L. Thus, we can express \(C\) in terms of the weight of insulin and its molecular weight.

### Step 3: Relate osmotic pressure to the slope of the graph
From the slope of the plot of π against C, we can express the slope (m) as:
\[ \text{slope} = \frac{1000RT}{M} \]
Given that the slope is \(4.65 \times 10^{-3}\), we can set up the equation:
\[ 4.65 \times 10^{-3} = \frac{1000RT}{M} \]

### Step 4: Convert temperature to Kelvin
The temperature given is \(20^\circ C\). To convert this to Kelvin:
\[ T(K) = 20 + 273 = 293 \, K \]

### Step 5: Substitute values into the equation
Now, substituting \(R = 0.0821 \, \text{L·atm/(K·mol)}\) and \(T = 293 \, K\) into the equation:
\[ 4.65 \times 10^{-3} = \frac{1000 \times 0.0821 \times 293}{M} \]

### Step 6: Solve for M (molecular weight)
Rearranging the equation to solve for \(M\):
\[ M = \frac{1000 \times 0.0821 \times 293}{4.65 \times 10^{-3}} \]

### Step 7: Calculate the molecular weight
Now, calculate \(M\):
\[ M = \frac{1000 \times 0.0821 \times 293}{4.65 \times 10^{-3}} \approx 5.17 \times 10^6 \, \text{g/mol} \]

### Final Answer
The molecular weight of insulin is approximately \(5.17 \times 10^6 \, \text{g/mol}\).

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