Answer:
b. 0.100 m Li₂SO₄
Explanation:
The elevation in the boiling point of solvents depends on the nature of the solute and its concentration. The formula for elevation is given as,
[tex]\Delta T_b=i\times K_b \times m[/tex]
Here ΔTb is the elevation in the boiling point, Kb is the boiling point elevation constant for a solvent and m represents the molality. i accounts for the number of species present in the solution; so that the ionic compounds that dissociate into multiple ions in water are also taken into consideration.
In the given problem solvent is water, hence the constant will remain the same while i and m values will vary with the type of solute. This means that taking the product of i and m value of each species, we can determine which solute will create the highest elevation in the boiling point of water
a) [tex]i_{NaNO_3} \times m_{NaNO_3}= 2 \times 0.100 = 0.2[/tex]
b) [tex]i_{Li_2SO_4} \times m_{Li_2SO_4}= 3 \times 0.100 = 0.3[/tex]
c) [tex]i_{C_3H_8O_4} \times m_{C_3H_8O_4}= 1 \times 0.200 =0.2[/tex]
d) [tex]i_{Na_3PO_4} \times m_{Na_3PO_4}= 4 \times 0.060=0.24[/tex]
The above calculations show that the boiling point elevation will be highest if the solute is Li₂SO₄.