Question

1. If 14.95 mL of a 0.1250 M HCl solution is added to reach the endpoint of a titration, how many moles of HCl have been added?

		0.001869 moles
		0.008361 moles
		8.361 moles
		0.1196 moles
		1.869 moles

QUESTION 2

A buret is read to how many decimal places?

		3
		can only be read to the ones place - so no decimal places
		2
		4
		1

QUESTION 3

As the endpoint nears what happens to the color?

		nothing can visibly be seen until the endpoint happens
		the color begins being more prominent, but still disappears
		the color is very vibrant but then will go away when the endpoint happens
		there is no color change yet
		color is perminent

  

QUESTION 4

A 20.00 mL sample of 0.1015 M nitric acid is introduced into a flask, and water is added until the volume of the solution reaches 250. mL. What is the concentration of nitric acid in the final solution?

		5.08 × 10–4 M
		3.25 × 10–2 M
		8.12 × 10–3 M
		0.406 M
		1.27 M

QUESTION 5

If 25.0 mL of an unknown concentration of sodium hydroxide is added to a flask and then titrated with 0.1025 M HCl. If 24.69 mL of the acid was added, what is the concentration of sodium hydroxide?

		63.27 M
		0.6741 M
		0.1012 M
		0.1038 M
		9.879 M

Answer 1

Ans 1: Correct option is 0.001869 moles

Explanation:

Number of moles of HCl = Volume of HCl (L) x Molarity of HCl solution

Number of moles of HCl = (14.95mL x 1L/1000mL) x 0.1250 M

Number of moles of HCl = 0.001869 mol

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Ans 2: Correct answer is 2.

Explanation:

The smallest gradation on burette is 0.1mL. However, second decimal can be determined depending upon where the bottom of miniscus lies between two 0.1 marks. Second decimal will be an estimated value.

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Ans 3: correct answer is the color begins being more prominent, but still disappears

Explanation:

As the end point nears, the color starts becoming more prominent but it disappears. At the end point, the color does not disappear.

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Ans 4: Correct answer is 8.12 × 10^–3 M

Explanation:

Number of moles of nitric acid = Volume (L) x Molarity

Number of moles of nitric acid = (20.00mL x 1L/1000mL) x 0.1015

Number of moles of nitric acid = 0.00203 mol

Thus, molarity of diluted solution = number of moles / volume(L)

molarity of diluted solution = (0.00203 mol) / 0.250L

molarity of diluted solution = 8.12 × 10^–3 M

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Ans 5: Correct answer is 0.1012 M

Explanation:

Moles of HCl required = Volume of HCl (L) x molarity of HCl

Moles of HCl required = (24.69mL x 1L/1000mL) x 0.1025M

Moles of HCl required = 0.002531 mol

Since, at the end point moles of HCl = moles of NaOH

Therefore, moles of NaOH = 0.002531 mol

Concentration of NaOH = Moles / volume (L)

Concentration of NaOH = 0.002531 mol / 0.025L

Concentration of NaOH = 0.1012M