A solution of glycerol `C_(3)H_(8)O_(3)` , molar mass = 92 g `mol^(-1)` in water was prepared by dissolving some glycerol 500 g of water. This solution has a boiling point of `100.42^(@)C`. What mass of glycerol was dissolved to make this solution ? `K_(b)` for water=` 0.512 K kg mol^(-1).`

Explanation

To find the mass of glycerol dissolved in 500 g of water, we can follow these steps:

### Step 1: Calculate the change in boiling point (∆Tb)
The boiling point of the solution is given as 100.42 °C, and the boiling point of pure water is 100 °C. Therefore, we can calculate ∆Tb as follows:

\[ \Delta T_b = T_{b, \text{solution}} - T_{b, \text{solvent}} = 100.42 °C - 100 °C = 0.42 °C \]
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