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Volumetric Analysis

Volumetric Analysis

Assessment

Presentation

Chemistry

11th - 12th Grade

Medium

NGSS
HS-PS1-7

Standards-aligned

Created by

Angela Toh

Used 5+ times

FREE Resource

8 Slides • 15 Questions

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Volumetric Analysis

By Angela Toh

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Multiple Choice

The point in an acid-alkali titration at which the reactants just react completely with each other is called the __________ point.

1

equivalence point

2

end point

4

Multiple Choice

A chemical used to show the end point of titration is called

1

base

2

indicator

3

pipette

4

acid

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Multiple Choice

The quantity of liquid delivered by the burette in a titration is called

1

burette reading

2

aliquot

3

concentration

4

titre

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Multiple Choice

To prepare a solution of accurately known volume, use a

1

measuring cylinder

2

beaker

3

concial flask

4

volumetric flask

8

Multiple Choice

A standard solution is a solution with accurately known concentration.

1

True

2

False

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Multiple Choice

What volume of water is required to dilute 120 cm3 of 10 mol dm–3 sulfuric acid to a concentration of 2 mol dm–3?

1

360 cm3

2

480 cm3

3

600 cm3

4

720 cm3

13

Multiple Choice

3.75 dm3 of distilled water are added to 250.0 cm3 of 3.20 mol dm–3 sodium hydroxide solution. What is the molarity of the diluted solution?

1

0.200 mol dm-3

2

0.213 mol dm-3

3

2.00 mol dm-3

4

2.13 mol dm-3

14

Multiple Choice

What volume of 5.0 mol dm–3 hydrochloric acid is required to prepare 250 cm3 of a 0.40 mol dm–3 solution?

1

10 cm3

2

20 cm3

3

30 cm3

4

40 cm3

15

Fill in the Blanks

Beaker A contains 500 cm3 of 1.0 M of HCl. A student pipettes 25 cm3 of HCl

from beaker A and places it in Beaker B. The student then adds water to Beaker B such that the total volume of solution in Beaker B is 250 cm3. What is the final concentration (in terms of M) of HCl in Beaker B?

Type answer...

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Multiple Choice

A 4 g sample of sodium hydroxide, NaOH, is dissolved in water and made up to 500

cm3 of aqueous solution. What is the concentration of the resulting solution?

1

0.1 mol dm-3

2

0.2 mol dm-3

3

0.5 mol dm-3

4

1 mol dm-3

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Open Ended

27.82 g of hydrated sodium carbonate crystals, Na2CO3. xH2O, was dissolved in water and made up to 1.000 dm3. 25.00 cm3 of this solution was neutralized by 48.80 cm3 of hydrochloric acid of concentration 0.1000 mol dm-3.

(a) Write an equation for the reaction between sodium carbonate and hydrochloric

acid, including the state symbols.

20

Fill in the Blanks

27.82 g of hydrated sodium carbonate crystals, Na2CO3. xH2O, was dissolved in water and made up to 1.000 dm3. 25.00 cm3 of this solution was neutralized by 48.80 cm3 of hydrochloric acid of concentration 0.1000 mol dm-3.

(b) Calculate the molar concentration of the sodium carbonate solution neutralized

by the hydrochloric acid.

Type answer...

21

Fill in the Blanks

27.82 g of hydrated sodium carbonate crystals, Na2CO3. xH2O, was dissolved in water and made up to 1.000 dm3. 25.00 cm3 of this solution was neutralized by 48.80 cm3 of hydrochloric acid of concentration 0.1000 mol dm-3.

(c) Determine the mass of sodium carbonate neutralized by the hydrochloric acid

and hence the mass of sodium carbonate present in the 1.000 dm3 of solution.

Type answer...

22

Fill in the Blanks

27.82 g of hydrated sodium carbonate crystals, Na2CO3. xH2O, was dissolved in water and made up to 1.000 dm3. 25.00 cm3 of this solution was neutralized by 48.80 cm3 of hydrochloric acid of concentration 0.1000 mol dm-3.

(d) Calculate the mass of water in the hydrated crystals.

Type answer...

23

Fill in the Blanks

27.82 g of hydrated sodium carbonate crystals, Na2CO3. xH2O, was dissolved in water and made up to 1.000 dm3. 25.00 cm3 of this solution was neutralized by 48.80 cm3 of hydrochloric acid of concentration 0.1000 mol dm-3.

(d) Use the mass of water in the hydrated crystals to find the value of

x.

Type answer...

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Volumetric Analysis

By Angela Toh

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