Question

. What is molarity of K+ in aqueous solution that contains
17.4 ppm of K2SO4 (molecular weight of K2SO4 is 174
g/mol.)​

Answer 1

The molarity of K⁺ in aqueous solution that contains 17.4ppm of K₂SO₄ is 2×10⁻⁴ M.

Explanation:

1.Molarity of a solution is defined as the no. of moles of solute dissolved per litre of the solution.Mathematically,
                       $Molarity=\frac{Number of moles of the solute}{Volume of the solution in litres}$
2. Again 1ppm=mass of solute present in 10⁶ parts by mass of solution
3.According to the question, 17.4 ppm of K₂SO₄ is present in the aqueous solution containing K⁺ ions.

                 ∴17.4ppm of K₂SO₄⇒17.4mg/L of K₂SO₄
                                                 ⇒17.4 ×10⁻³gm of K₂SO₄   [∵1mg=10⁻³gm]
4.Therefore, the number of moles of K₂SO₄ could be found out to be
                            $Number of moles=\frac{weight}{molecular weight}$
                                                        =17.4×10⁻³gm
                                                           174gm/mol
                                                        =10⁻⁴moles of K₂SO₄
5. Now K₂SO₄ dissociates as,
                            K₂SO₄⇄ 2K⁺+ SO₄⁻
      ∴Number of moles of K⁺ = 2×10⁻⁴
6.Hence molarity of K⁺=2×10⁻⁴moles
                                               1L
                                    =2×10⁻⁴M
Therfore molarity of K⁺ ion is 2×10⁴M.

Answer 2

Solution:

1 ppm = 1mg/L

$1ppm = 10^{-3 } g/L$

17.4 ppm of $K_2SO_4$ solution = $17.3 \times 10^{-3 } g/L$ solution

Weight of  $K_2SO_4$ in 1 Litre solution = $17.3 \times 10^{-3 } g$

Molecular weight of $K_2SO_4$ = 174 g/mol

 $\text{Number of moles of} \ K_2SO_4 = \frac{\text {given weight}}{\text {molecular weight}}$

$\text{Number of moles of} \ K_2SO_4 = \frac{17.4\times 10^{-3}}{174}$

$\text{Number of moles of} \ K_2SO_4 = 10^{-4} moles$

$K_2SO_4$ dissociates in solution as:

$K_2SO_4\rightleftharpoons 2K^{+} + SO_4 ^{2-}$

1 mole of  $K_2SO_4$ dissociates to produce 2 moles of $K^{+} ions$

$10^{-4} moles$ of  $K_2SO_4$ dissociates to produce = $2\times 10^{-4} moles$ of $K^{+} ions$

$Molarity = \dfrac{\text{Number of moles}}{\text{Volume of solution in L}}$

$\text{Molarity of} \ K^+ = \dfrac{2\times 10^{-4}}{1}$

$\text{Molarity of} \ K^+$ in aqueous solution that contains 17.4 ppm of $K_2SO_4$ is ${2\times 10^{-4} mol/L$