(a) `A` solution is prepared by dissolving `10g` of non-volatile solute in `180g` of `H_(2)O`. If the relative lowering of vapour pressure is `0.005`, find the molar mass of the solute.
(b) How many grams of sucrose `(C_(12)H_(22)O_(11))`must be dissolved in `90g` of water to produce a solution over which the relative humidity is `80%`? Assume the solution is ideal.

Explanation

Let's solve the problem step by step.

### Part (a)

**Step 1: Understanding Relative Lowering of Vapor Pressure**
The relative lowering of vapor pressure (ΔP/P₀) is equal to the mole fraction of the solute (X_solute). Given that the relative lowering of vapor pressure is 0.005, we can write:
\[ \Delta P/P_0 = 0.005 \]
Thus,
\[ X_{\text{solute}} = 0.005 \]

**Step 2: Expressing Mole Fraction**
The mole fraction of the solute can be expressed as:
\[ X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \]
Where:
- \( n_{\text{solute}} = \frac{m_{\text{solute}}}{M_{\text{solute}}} \)
- \( n_{\text{solvent}} = \frac{m_{\text{solvent}}}{M_{\text{solvent}}} \)

**Step 3: Calculating Moles of Solvent**
The mass of the solvent (water) is given as 180 g. The molar mass of water (H₂O) is 18 g/mol. Therefore, the number of moles of the solvent is:
\[ n_{\text{solvent}} = \frac{180 \text{ g}}{18 \text{ g/mol}} = 10 \text{ moles} \]

**Step 4: Setting Up the Equation**
Substituting the values into the mole fraction equation:
\[ 0.005 = \frac{\frac{10 \text{ g}}{M_{\text{solute}}}}{\frac{10 \text{ g}}{M_{\text{solute}}} + 10} \]
Let \( M_{\text{solute}} \) be the molar mass of the solute.

**Step 5: Simplifying the Equation**
This simplifies to:
\[ 0.005 = \frac{10}{M_{\text{solute}} + 10M_{\text{solute}}} \]
Cross-multiplying gives:
\[ 0.005(M_{\text{solute}} + 10) = 10 \]
Expanding this:
\[ 0.005M_{\text{solute}} + 0.05 = 10 \]
Rearranging gives:
\[ 0.005M_{\text{solute}} = 10 - 0.05 = 9.95 \]

**Step 6: Solving for Molar Mass**
\[ M_{\text{solute}} = \frac{9.95}{0.005} = 1990 \text{ g/mol} \]

### Answer for Part (a)
The molar mass of the solute is **1990 g/mol**.

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### Part (b)

**Step 1: Understanding Relative Humidity**
Given that the relative humidity is 80%, we can express the new vapor pressure (P) as:
\[ \frac{P}{P_0} = 0.8 \implies P = 0.8P_0 \]

**Step 2: Finding Relative Lowering of Vapor Pressure**
The relative lowering of vapor pressure is given by:
\[ \frac{P_0 - P}{P_0} = \frac{P_0 - 0.8P_0}{P_0} = 0.2 \]
Thus, the relative lowering of vapor pressure is 0.2.

**Step 3: Setting Up the Mole Fraction Equation**
Using the mole fraction of the solute:
\[ 0.2 = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \]
Where:
- \( n_{\text{solute}} = \frac{x}{342} \) (where x is the mass of sucrose)
- \( n_{\text{solvent}} = \frac{90 \text{ g}}{18 \text{ g/mol}} = 5 \text{ moles} \)

**Step 4: Substituting Values**
Substituting into the equation:
\[ 0.2 = \frac{\frac{x}{342}}{\frac{x}{342} + 5} \]

**Step 5: Cross-Multiplying**
Cross-multiplying gives:
\[ 0.2\left(\frac{x}{342} + 5\right) = \frac{x}{342} \]
Expanding this:
\[ 0.2\frac{x}{342} + 1 = \frac{x}{342} \]
Rearranging gives:
\[ 1 = \frac{x}{342} - 0.2\frac{x}{342} \]
\[ 1 = 0.8\frac{x}{342} \]

**Step 6: Solving for x**
\[ x = \frac{342}{0.8} = 427.5 \text{ g} \]

### Answer for Part (b)
To produce a solution over which the relative humidity is 80%, **427.5 g** of sucrose must be dissolved in 90 g of water.

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