Model Question Paper — 2 — Question 19

(i) Given,

solute = 20 g

solvent = 200 g

concentration = ?

Concentration of solution = $\dfrac{\text{Mass of solute}}{\text{\text{Mass of solute + Mass of solvent}}}$ x 100

Concentration of solution = $\dfrac{20}{\text{200 + 20}}$ x 100

= 9.09%

Hence, concentration = 9.09%

(ii) Given,

concentration = 25%

solvent = 150 cm³

solute = ?

Concentration of solution = $\dfrac{\text{Volume of solute}}{\text{\text{Volume of solute + Volume of solvent}}}$ x 100

$\dfrac{25}{100}$ = $\dfrac{\text{Volume of solute (x)}}{\text{\text{Volume of solute (x)+ 150 }}}$

$\dfrac{1}{4}$ = $\dfrac{\text{Volume of solute (x)}}{\text{\text{Volume of solute (x)+ 150 }}}$

4x = x + 150

3x = 150

x = 50 cm³

Hence, alcohol added = 50 cm³

(iii) Solubility of a substance at 45°C is 36.5g means that 36.5g of substance dissolves in 100 g of water at the temperature of 45°C.