Question-
Question- The concentration of Mg2• in the solution made by mixing 10 mL of 0.25 M 25 ml- of 0.2 M NaF will be = 8 q 10") (A) 00027 M (B) 00714 M (C) 0.0030 M (D 0.0060 M The pH Of a saturated solution of in water will be — 4 W 12)

Dear Student,

Mg ({NO}_{3})_{2} + 2 NaF \to {MgF}_{2} + 2 {NaNO}_{3} 1 mol 2 mol 1 mol Moles of Mg ({NO}_{3})_{2} = 0.01 L \times 0.25 M = 0.0025 mol Moles of NaF = 0.025 L \times 0.2 M = 0.005 mol Mg ({NO}_{3})_{2} is the limiting reagent here as it is less and will be consumed first . So, 1 mol of Mg ({NO}_{3})_{2} gives 1 mol of {MgF}_{2} So, 0.0025 mol of Mg ({NO}_{3})_{2} gives 0.0025 mol of {MgF}_{2} and, moles of {Mg}^{2 +} = moles of {MgF}_{2} and, [{Mg}^{2 +}] = \frac{moles of {MgF}_{2}}{Total volume of solution} = \frac{0.0025}{0.01 + 0.025} = 0.0714 M Hence, correct answer is (B)

​​Question-​​​Question-​ The concentration of Mg2• in the solution made by mixing 10 mL of 0.25 M 25 ml- of 0.2 M NaF will be = 8 q 10") (A) 00027 M (B) 00714 M (C) 0.0030 M (D 0.0060 M The pH Of a saturated solution of in water will be — 4 W 12)