What is the molarity of K +in aqueous solution that contains 17.4 ppm of K 2 SO 4 (molar mass =174 g mol -1 )?

Question

What is the molarity of K +in aqueous solution that contains 17.4 ppm of K 2 SO 4 (molar mass =174 g mol -1 )? (A) 2 × 10-2 M (B) 2 × 10-3 M (C) 4 × 10-4 M (D) 2 × 10-4 M

Tardigrade Admin · Accepted Answer

K 2 SO 4 is 17.4 ppm, i.e. 106 g has 17.4 g K 2 SO 4 1 L (103 mL ) has K 2 SO 4 =(17.4 × 103/106)=0.0174 g / L =(0.0174/174) mol / L ∴ [ K 2 SO 4]=1 × 10-4 M K 2 SO 4 leftharpoons 2 K ++ SO 42- ∴ [ K +]=2 × 10-4 M

Q. What is the molarity of $K^{+}$ in aqueous solution that contains $17.4 pp m$ of $K_{2} S O_{4}$ (molar mass $= 174 g m o l^{- 1}$ )?

$2 \times 1 0^{- 2} M$

$2 \times 1 0^{- 3} M$

$4 \times 1 0^{- 4} M$

$2 \times 1 0^{- 4} M$

Q. What is the molarity of K+in aqueous solution that contains 17.4ppm of K2​SO4​ (molar mass =174gmol−1 )?

2×10−2M

2×10−3M

4×10−4M

2×10−4M

K2​SO4​ is 17.4ppm, i.e. 106g has 17.4gK2​SO4​ 1L(103mL) has K2​SO4​ =10617.4×103​=0.0174g/L =1740.0174​mol/L ∴[K2​SO4​]=1×10−4M K2​SO4​⇌2K++SO42−​ ∴[K+]=2×10−4M

Q. What is the molarity of $K^{+}$ in aqueous solution that contains $17.4 pp m$ of $K_{2} S O_{4}$ (molar mass $= 174 g m o l^{- 1}$ )?

$2 \times 1 0^{- 2} M$

$2 \times 1 0^{- 3} M$

$4 \times 1 0^{- 4} M$

$2 \times 1 0^{- 4} M$