Mass percentage (w / w) =  \(\frac{\text { Mass of the solute }}{\text { Mass of the solution }} \times 100\)
\(10=\frac{\text { Mass of the urea }}{500} \times 100\)
Mass of urea = 50 g

(i)  Mass percentage of the solute
Mass percentage of solute = \(\begin{aligned} & \frac{\text { Mass of the solute }}{\text { Mass of the solute +Mass of the solvent}} \times 100 \\ \end{aligned}\)
= \(\frac{16 × 100}{16 + 100}\)
\(=\frac{1600}{116}\)
Mass percentage of the solute = 13.79 %
(ii) Mass percentage of solvent = 100 - (Mass percentage of the solute)
= 100 – 13.79
= 86.21%

Mass of the solute = 25 g
Mass of the solvent = 100 g
\(\begin{aligned} \text { Mass Percentage } \end{aligned}=\frac{\text { Mass of the solute }}{\text { Mass of the solution }} \times 100\)
\(\begin{aligned} \text { Mass Percentage } \end{aligned}=\frac{\text { Mass of the solute }}{\begin{array}{l} \text { Mass of the solute }+ \text { Mass of the solvent } \end{array}} \times 100\)
\(=\frac {25}{25+100}\times\)100
= \(\frac{25}{125}\times\)100
= 20%

Amount of sodium nitrate dissolved in 100 g of water at 50°C is 114 g
\(\therefore\) Amount of sodium nitrate dissolving in 50 g of water at 50°C is = \(\frac{114\times50}{100}\)
= 57 g
Similarly amount of sodium nitrate dissolving in 50g of water at 30°C is = \(\frac{96 \times 50}{100}\)
= 48 g
Amount of sodium nitrate thrown when 50g of water is cooled from 50°C to 30°C is 
57 – 48 = 9 g

Mass of the solute = 1.5 g
Mass of the solvent = 15 g
Solubility of the solute = \(\frac{\text { Mass of the solute }}{\text { Mass of the solvent }} \times 100\) 
Solubility of the solute = \(\frac{1.5}{15}\times 100\) 
= 10 g

Given, Volume of ethanol = 3.5 lit
Volume of aqueous ethanol solution = 15 lit
Volume percent ethanol solution = ?
Volume percentage = \(\frac {Volume \ of \ the \ solute}{Volume \ of \ the \ solution}\) x 100
= \(\frac{3.5}{15}\times 100=\frac{350}{15}\) 
Volume percentage of ethanol solution = 23.33%

mass of solute = 45 g
mass of solvent = 180 g
Mass of Percentage = ?
Mass of percentage = \(\frac{Mass \ of \ the \ solute}{Mass \ of \ the \ solution \ + \ Mass \ of \ the \ solvent}\) x 100
= \(\frac{45}{45 + 180} \times 100=\frac{4500}{225}\)
Mass percentage = 20%

a) MgSO₄.7H₂O is heating process:
(i) Magnesium Sulphate heptahydrate or Epsom salt MgSO₄.7H₂O water of crystallization is 7.
(ii) When magnesium sulphate heptahydrate crystals are gently heated, it loses seven water molecules, and becomes anhydrous magnesium sulphate.

(iii) If you add few drops of water or allow it to cool, the colourless anhydrous salt again turns back into hydrated salt.
b) Solubility:
b) Solubility is defined as the number of grams of a solute that can be dissolved in 100g of a solvent to form its saturated solution at a given temperature and pressure.
Solubility = \(\frac{Mass \ of \ the \ solute}{Mass \ of \ the \ solvent}\) x 100

There are three main factors which govern the solubility of a solute. They are:
a) Nature of the solute and solvent
b) Temperature
c) Pressure
a) Nature of the solute and solvent:
(i) The nature of the solute and solvent plays an important role in solubility.
(ii) Although water dissolves an enormous variety of substances, both ionic and covalent, it does not dissolve everything.
(iii) The phrase that scientists often use when predicting solubility is "like dissolves like".
(iv) The expression means that dissolving occurs when similarities exist between the solvent and the solute.
(v) For example: Common salt is a polar compound 4 and dissolves readily in polar solvent like water.
(vi) Non-polar compounds are soluble in non-polar solvents. For example: Fat dissolved in ether.
(vii) But non-polar compounds, do not dissolve in polar solvents; polar compounds do not dissolve in non-polar solvents.
b) Effect of Temperature:
Solubility of Solids in Liquid:
(i) Generally, solubility of a solid solute in a liquid solvent increases with increase in temperature.
(ii) For example, a greater amount of sugar will dissolve in warm water than in cold water.
(iii) In endothermic process, solubility increases with increase in temperature.
(iv) In exothermic process, solubility decreases with increase in temperature.
Solubility of Gases in liquid:
(i) Solubility of gases in liquid decrease with increase in temperature.
(ii) Generally, water contains dissolved oxygen.
(iii) When water is boiled, the solubility of oxygen in water decreases, so oxygen escapes in the form of bubbles.
(iv) Aquatic animals live more in cold regions because, more amount of dissolved oxygen is present in the water of cold regions.
(v) This shows that the solubility of oxygen in water is more at low temperatures.
c) Effect of Pressure:
(i) Effect of pressure is observed only in the case of solubility of a gas in a liquid.
(ii) When the pressure is increased, the solubility of a gas in liquid increases.
(iii) The common examples for solubility of gases in liquids are carbonated beverages, i.e. soft drinks, household cleaners containing aqueous solution of ammonia, formalin aqueous solution of formaldehyde, etc.

(i) Saturated solution:
a) A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution.
b) E.g. 36 g of sodium chloride in 100 g of water at 25° C forms saturated solution.
c) Further addition of sodium chloride, leaves it undissolved.
(ii) Unsaturated solution:
a) Unsaturated solution is one that contains less solute than that of the saturated solution at a given temperature.
b) E.g. 10 g or 20 g or 30 g of Sodium chloride in 100 g of water at 25°C forms an unsaturated solution.