22. Volumetric methods involve the measurement of the (B) volume of the titrant

Question

22. Volumetric methods involve the measurement of the (B) volume of the titrant (A) density (C) mass (D) temperature 23. How do you prepare 250.0 mL of 0.500 M glucose standard solution starting with 5.00 M glucose solution? (A) Mix 25.0 mL of 5.00 M glucose solution with 225 mL distilled water in a beaker. (B) Dilute 100 mL of 5.00 M glucose solution with distilled water to the mark using a 250.0 mL volumetric flask (C) Dilute 25.0 mL of 5.00 M glucose solution with distilled water to the mark using a 250.0 mL volumetric flask. (D) Mix 25.0 mL of 5.00 M glucose solution with 250 mL distilled water in a beaker. 24. Which of the following is a suitab le primary standard that can be used to standardize an HCI solution? (A) sodium carbonate (C) potassium hydrogen phthalate (B) sodium oxalate (D)potassium permanganate 25. How many moles of BrO; are present in 0.1262 g of high purity KBrO,? [K-39.0983, Br-79.904, O-15.9994] (A) 7.556 x 10 mol (C) 1.323 x 10 mol (B) 9.858 x 10 mol (D) 21.08 mol 26. A volumetric analysis required 35.20 mL of 02105 M NaOH to titrate a sample ofa - tic acid with a molar mass of 140.1 gmol. Calculate the mass of acid present in this sample. HA(aq) + NaOH(a) (A) 0.9633 g (B) 1.038 g NaA(aq) H2O(D (C) 2.610g (D) 2.945 g 27. A 0.3708 g sample of a monoprotic weak acid was dissolved in water and titrated with 0.1125 M NaOH solution. The volume of NaOH solution required to reach the end point was 34.05 mL. What is the molar mass of the weak acid? (A) 96.80 g/mol (C) 142.1 g/mol (B) 104.1 g/mol (D) 214.2 g'mol 28. A 0.3258-g sample of primary standard Na COs (molar mass- 105.99 g/mol) required 24.65 mL of HCl solution to reach the end point. Determine the molarity of HCl solution. Na COs + 2 HC (A) 0.2494 M (C) 0.4010 M 2 NaCl +CO2+ H2O (B) 0.3690 M (D) 0.4988 M

Explanation / Answer

THE ANSWERS:

22. B

23. A - use M1V1 = M2 V2

24. A

25. A - no.of moles = given mass/ molar mass

26. B - 35..2 ml = 0.0352 l, no. of moles of NaOH= 0.0352 x 0.2105 =0.00740 = no. of moles of acid ( according to equation), mass of acid = no.of moles x molar mass = 0.0074 moles x 140 g/mol = 1.038 g

27. A - use the above formula

28. A - use molarity = no.of moles of solute/ solution in litre, no. of moles of HCl = 2 x no. of moles of NaOH

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.