Question

6. The mass of saturated solution of urea in water is 30.7 g. The
solution is evaporated to dryness. The mass of residue left is 3.2 g.
Calculate the solubility of urea in water at that temperature.​

Answer 1

Answer:

Solubility product of CaCO₃ is 4.89 x 10⁻³.

Explanation:

Solubility product of CaCO₃ is 4.89 x 10⁻³.

Answer 2

Answer:

The solubility of urea in water at that temperature.​ 4.9x 10 power negative 3.

Explanation:

Given: The mass of saturated solution of urea in water is 30.7 g.

            The mass of residue left is 3.2 g.

To find: the solubility of urea in water at that temperature.​

• CaCO3 - ca+2+CO2-3
• Molar mass of CaCO3= 100
• $[Ca^{+2} ]=\frac{\frac{7g}{100} }{1 litre} \\=0.07$
• =4.9x 10 power negative 3.
• Urea dissolves in water very easily. it is very easily soluble in its formation with water.
• The specific gravity of urea is 1.34 at room temperature:
• 68 degrees Fahrenheit or 20 degrees Celsius.

Hence, the solubility of urea in water at that temperature.​ 4.9x 10 power negative 3.