Quantitative Analysis of Organic Compounds Question And Answers – NEET General Organic Chemistry

Quantitative Analysis Of Organic Compounds

Question 1. The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed into 100 mL of 0.1 M sulphuric acid. The excess of acid requires 20 mL of 0.5 M sodium hydroxide solution for complete neutralization, the organic compound is

1. Acetamide
2. Benzamide
3. Urea
4. Thiourea

Answer: 3. Urea

Solution:

Let unreacted 0.1 M(= 0.2 N) H₂SO₄ = V mL

20ml of 0.5 NaOH = V mL of 0.2 N H₂SO₄

⇒ V= \(\frac{20 \times 0.5}{0.2}\)

= 50 mL

20 x 0.5 = V x 0.2

Used H2SO4 = 100 – 50

= 50mL

⇒ \(\% \text { of } N=\frac{1.4 N V}{w}=\frac{1.4 \times 0.2 \times 50}{0.30}=46.67 \%\)

% Of nitrogen in

⇒ 1. \(\mathrm{CH}_3 \mathrm{CONH}_2=\frac{14 \times 100}{59}=23.73 \%\)

⇒ 2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CONH}_2=\frac{14 \times 100}{121}=11.57 \%\)

⇒ 3. \(\mathrm{NH}_2 \mathrm{CONH}_2=\frac{28 \times 100}{60}=46.67 \%\)

⇒ 4. \(\mathrm{NH}_2 \mathrm{CSNH}_2=\frac{28 \times 100}{76}=36.84 \%\)

Read And Learn More: NEET General Organic Chemistry Notes, Question And Answers

Question 2. The simplest formula of a compound containing 50% of element X (at. wt 10) and 50% of element Y (at. wt. 20) is

1. XY
2. XY₂
3. X₂Y
4. X₂Y₂

Answer: 3. X₂Y

Solution:

Hence, the empirical formula = X₂Y

Question 3. 0.25 g of an organic compound on Kjeldahl’s analysis gave enough ammonia to just neutralize 10cm³ of 0.5M H₂SO₄. The percentage of nitrogen in the compound is

1. 28
2. 56
3. 14
4. 112

Answer: 2. 56

Solution:

⇒ \(\text { Percent of nitrogen }=\frac{1.4 \times N \times V}{W}\)

⇒ \(=\frac{1.4 \times 0.5 \times 2 \times 10}{0.25}=56 \%\)

0.25 g of an organic compound on Kjeldahl’s analysis gave enough ammonia to just neutralize 10cm³ of 0.5M H₂SO₄. The percentage of nitrogen in the compound is 56

Question 4. What is the empirical formula of a compound having 40% carbon, 6.66% hydrogen, and 53.34% oxygen?

1. C₂H₂O
2. C₂H₄O
3. CH₂O
4. CHO

Answer: 3. CH₂O

Solution:

Hence, empirical formula = CH₂O

Question 5. Duma’s method involves the determination of nitrogen content in the organic compound in the form of

1. NH₃
2. CH₂O
3. CN
4. (NH₄)₂SO₂

Answer: 2. CH₂O

Solution: Duma’s method involves the determination of nitrogen content in the organic compound in the form of N_2.

N₂O + Cu → N₂ + Cu

⇒ \(d \% \text { of } N=\frac{28}{22400} \times \frac{\text { volume of } N_2 \text { at } N T P}{\text { weight of compound }} \times 100\)

Question 6. 0.4 g of a silver salt of a monobasic organic acid gave 0.26 g pure silver on ignition. The molecular weight of the acid is (atomic weight of silver=108)

1. 58
2. 37
3. 89
4. 105

Answer: 1. 58

Solution:

Mass of silver salt  = 0.4 g

Mass of silver = 0.26 g

⇒ Eq. mass of silver salt/Eq.mass of Ag = \(\frac{\text { wt.of silver salt }}{\text { wt.of silver }}\)

⇒ Eq.mass of silver salt = \(\frac{108 \times 0.4}{0.26}\)

= 166

Eq.mass of acid = 166-108

= 58

0.4 g of a silver salt of a monobasic organic acid gave 0.26 g pure silver on ignition. The molecular weight of the acid is 58

Question 7. An organic compound contains 49.3% carbon, 6.84% hydrogen, and 43.86% oxygen, and its vapor density is about 73. The molecular formula of the compound is

1. C₆H₉O₃
2. C₂H₁₀O₂
3. C₂H₅O₂
4. C₃H₁₀O₂

Answer: 1. C₆H₉O₃

Solution:

So, formula=(C₂H₃O)₃≈C₆H₉O₃

An organic compound contains 49.3% carbon, 6.84% hydrogen, and 43.86% oxygen, and its vapor density is about 73. The molecular formula of the compound is C₆H₉O₃

Question 8. Quantitative measurement of nitrogen in an organic compound is done by the method?

1. Berthelot method
2. Bilstein meth
3. Lassaigne’s test
4. Kjeldahl’s method

Answer: 4. Kjeldahl’s method

Solution: Kjeldahl’s and Duma’s methods are used for the quantitative estimation of nitrogen in an organic compound. In Kjeldahl’s method, the nitrogen element present in the organic compounds is changed to ammonia.

Question 9. In Kjeldahl’s method, the nitrogen present in the organic compound is quantitatively converted into

1. Gaseous ammonia
2. Ammonium sulfate
3. Ammonium phosphate
4. Ammonia

Answer: 4. Ammonia

Solution: In Kjeldahl’s method, the nitrogen is quantitatively converted into ammonia by heating with NaOH

C+H+N ∆ → (NH₄)SO₄+CO₂+H₂O

(From organic compound)

(NH₄)₂SO₄+2NaOH→ Na₂SO₄+2NH₃+2H₂O

In Dumas method nitrogen present in the organic compound is quantitatively converted into N_2..

Question 10. In Kjeldahl’s method of estimation of nitrogen, CuSO₄ acts as

1. Oxidizing agent
2. Reducing agent
3. Catalytic agent
4. Hydrolysis agent

Answer: 3. Catalytic agent

Solution: Kjeldahl’s method is used for the estimation of nitrogen. The organic compound is heated with conc. H₂SO₄ in the presence of K₂SO₄ (used to elevate the boiling point of H₂SO₄) and CuSO₄ (used as a catalyst) to convert all the nitrogen into (NH₄)₂SO₄.

Question 11. Phosphorus is estimated as

1. Na₃PO₄
2. P₂O₅
3. P₂O₃
4. Mg₂P₂O₇

Answer: 4. Mg₂P₂O₇

Solution: Phosphorous is estimated as Mg₂P₂O₇

P HNO₃ →_Δ H₃PO₄

H₃PO+ Mg²⁺

NH₂OH → MgNH₄PO₄

2MgNH₄PO₄ → Mg₂P₂O₇ + H₂O + 2NH₃

⇒ \(\% \text { Of P }=\frac{62 \times w t . o f ~ M g_2 P_2 O_7 \times 100}{222 \times w}\)

Question 12. In Kjeldahl’s method for the estimation of nitrogen, the formula used is

1. \(\% \text { of } N=\frac{1.4 \mathrm{Vw}}{N}\)
2. \(\% \text { of } N=\frac{1.4 \mathrm{VN}}{w}\)
3. \(\% \text { of } N=\frac{V N w}{1.8}\)
4. \(\% \text { of } N=\frac{1.4 w N}{V}\)

Answer: 2. \(\% \text { of } N=\frac{1.4 \mathrm{VN}}{w}\)

Solution: Fact.

Question 13.0.5 g of hydrocarbon gave 0.9 g water on combustion. The percentage of carbon in the hydrocarbon is

1. 60.6
2. 28.8
3. 80.0
4. 68.6

Answer: 3. 80.0

Solution:

⇒ \(\% \text { of } H=\frac{2}{18} \times \frac{\text { weight of } \mathrm{H}_2 \mathrm{O}}{\text { weight of organic compound }} \times 100\)

⇒ \(\frac{2}{18} \times \frac{0.9}{0.5} \times 100=20 \%\)

Question 14. The silver salt of a monobasic acid on ignition gave 60% of Ag. The molecular weight of the acid is

1. 37
2. 57
3. 73
4. 88

Answer: 3. 73

Solution:

⇒ \(\frac{E}{108}=\frac{100}{60}\)

⇒ Eq.wt. of the silver salt E = 108 x \(108 \times \frac{100}{60}=180\)

∴ Wq.wt. of the acid = E -108 + 1

= 73

Question 15. In the estimation of nitrogen by Duma’s method, 1.18 g of an organic compound gave 224 mL of N₂ at NTP. The percentage of nitrogen in the compound is

1. 20.0
2. 11.8
3. 47.7
4. 23.7

Answer: 4. 23.7

Solution:

⇒ \(\% \text { of } N \frac{28}{22400} \times \frac{\text { volume of } N_2 \text { at NTP }}{\text { wt.of compound }} \times 100\)

⇒  \(\frac{28}{22400} \times \frac{224}{1.18} \times 100\)

⇒ \(\frac{28}{1.18}\)

= 23.728

Question 16. 0.765g of acid gives 0.535g of CO₂ and 0.138 g of H₂O. Then, the ratio of the percentage of carbon and hydrogen is

1. 19:2
2. 18:11
3. 20:17
4. 1:7

Answer: 1. 19:2

Solution:

⇒ \(\% \text { of } \mathrm{C}=\frac{12}{44} \times \frac{0.535}{0.765} \times 100=19.07\)

⇒ \(\% \text { of } H=\frac{2}{18} \times \frac{0.138}{0.765} \times 100=2.004\)

Ratio of % of C: H=19:2 (approx.)

Question 17. An organic compound having carbon, hydrogen, and sulfur contains 4% of sulfur. The minimum molecular weight of the compound is

1. 500
2. 800
3. 400
4. 100

Answer: 2. 800

Solution:

As the min mol wt. must have at least one S-atom so

C = 10.5 g = \(\frac{10.5}{12} \mathrm{~mol}=0.87 \mathrm{~mol}\)

⇒ \(H=1 \mathrm{~g}=\frac{1}{1} \mathrm{~mol}=1 \mathrm{~mol}\)

⇒ \(\left(C_{0.87} H_1\right)_7=C_{6.09} H_7 \approx C_6 H_7\)

⇒ \(p V=n R T=\frac{w}{m} R T\)

⇒ \(\begin{gathered}
1 \times 1=\frac{2.4}{m} \times 0.082 \times 400 \\
m=79
\end{gathered}\)

Question 18. Mark the incorrect statement in the estimation of nitrogen by Kjeldahl’s method

1. Nitrogen gas is collected over caustic potash solution
2. Potassium sulfate is used as the boiling point elevator of H₂SO₄
3. Copper sulfate or mercury acts as a catalyst
4. Nitrogen is quantitatively decomposed to give ammonium sulfate

Answer: 1. Nitrogen gas is collected over caustic potash solution

Solution: In the case of Kjeldahl’s method, the percentage of N2 is calculated from the amount of NH₃

Question 19. A hydrocarbon contains 10.5 g carbon and 1 g hydrogen. Its 2.4 g mass has 1 L volume at 1 atm and 1270C. Hydrocarbon is

1. C₆H₇
2. C₆H₈
3. C₅H₆
4. C₆H₆

Answer: 1. C₆H₇

Solution:

⇒ \(C=10.5 \mathrm{~g}=\frac{10.5}{12} \mathrm{~mol}=0.87 \mathrm{~mol}\)

⇒ \(H=1 \mathrm{~g}=\frac{1}{1} \mathrm{~mol}=1 \mathrm{~mol}\)

⇒ ∴\(\left(C_{0.87} H_1\right)_7=C_{6.09} H_7 \approx C_6 H_7\)

⇒ \(p V=n R T=\frac{w}{m} R T\)

⇒ \(\begin{gathered}
1 \times 1=\frac{2.4}{m} \times 0.082 \times 400 \\
m=79
\end{gathered}\)

Hence, the hydrocarbon is C₂H₇

Question 20. 29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s method and the evolved ammonia was absorbed in 20mL of 0.1 M HCl solution. The excess of the acid required 15mL of 0.1M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is

1. 59.0
2. 47.4
3. 23.7
4. 29.5

Answer: 3. 23.7

Solution: Weight of organic compound = 29.5mg

NH₃+HCl → Na₄Cl

⇒ \(\mathrm{HCl} \underset{\text { (remaining) }}{+} \underset{\mathrm{NaOH}}{\longrightarrow} \underset{\mathrm{NaCl}}{\longrightarrow} \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}\)

⇒ Weight of nitrogen = \(\frac{14}{17} \times 8.5 \mathrm{mg}=7 \mathrm{mg}\)

⇒ % Nitrogen = \(\frac{7}{29.5} \times 1100=23.7 \%\)

Question 21. Carbon and hydrogen in organic compounds are estimated by

1. Kjeldahl’s method
2. Duma’s method
3. Liebig’s method
4. Carius method

Answer: 3. Liebig’s method

Solution: Carbon and hydrogen are estimated in organic compounds by Liebig’s method

C+2CuO∆→ 2Cu+CO₂

2H+CuO∆→ Cu+H₂O

The percentage of carbon and hydrogen is calculated from the weight of CO₂ and H₂ produced

Question 22. In the Carius method, 0.099 g of an organic compound gave 0.287 g AgCl. The percentage of chlorine in the compound will be

1. 28.6
2. 71.7
3. 35.4
4. 64.2

Answer: 2. 71.7

Solution:

⇒ % of chlorine = \(\frac{35.5}{143.5} \times \frac{\text { mass of } \mathrm{AgCl}}{\text { mass of the compound }} \times 100\)

⇒  \(\frac{35.5}{143.5} \times \frac{0.287}{0.099} \times 100\)

= 71.71 %

Question 23. An organic compound contains 29.27% carbon, 5.69 % hydrogen, and 65.04% bromine. Its empirical formula is

1. C₃H₅Br
2. C₃H₃Br
3. C₂H₄Br₂
4. C₃H7Br

Answer: 4. C₃H7Br

Solution:

⇒ \(C: H: B r=\frac{2.27}{12}: \frac{5.69}{1}: \frac{65.04}{80}\)

= 2.43:5.69:0.813

= 3:7:1

or empirical formula =C₃H₇Br

Question 24. In Kjeldahl’s method, ammonia from 5g of food neutralizes 30 cm3 of 0.1 N acid. The percentage of nitrogen in the food is

1. 0.84
2. 8.4
3. 16.8
4. 1.68

Answer: 1. 0.84

Solution: From Kjeldahl’s method, the Percentage of nitrogen

⇒ \(\begin{gathered}
=\frac{1.4 \times N \times V}{W}=\frac{1.4 \times 0.1 \times 30}{5} \\
=0.84 \%
\end{gathered}\)

Question 25. Incorrect statements among the following is

1. Aniline can be purified by steam distillation
2. The Bilstein test is not given by fluorine
3. Kjeldahl’s method is used for the estimation of sulfur
4. Lassaigne’s test is used in the qualitative detection of elements in organic compounds

Answer: 3. Kjeldahl’s method is used for the estimation of sulfur

Solution: Kjeldahl’s method is used for the estimation of nitrogen

Question 26. How much sulfur is present in an organic compound, if 0.53g of the compound gave 1.158g of BaSO4on analysis?

1. 10%
2. 15%
3. 20%
4. 30%

Answer: 4. 30%

Solution:

⇒ \(\% \text { of } S=\frac{32}{233} \times \frac{\text { wt.of } \mathrm{BaSO}_4}{\text { wt.of organic compound }} \times 100\)

⇒ \(\begin{gathered}
=\frac{32}{233} \times \frac{1.158}{0.53} \times 100 \\
=30 \%
\end{gathered}\)

Question 27. The presence of halogen in organic compounds can be detected using

1. Liebig’s test
2. Duma’s test
3. Kjeldahl’s test
4. Bilstein’s test

Answer: 4. Bilstein’s test

Solution: The presence of halogen in organic compounds can be detected by Bilstein’s test.

Question 28. An organic compound has carbon and hydrogen percentages in the ratio of 6:1 and carbon and oxygen percentages in the ratio of 3:4. The compound has the empirical formula

1. C₂H₆O
2. CHO₂
3. CH₄O
4. CH₂O

Answer: 4. CH₂O

Solution: C:H: O=6:1:8

⇒  \(\frac{6}{15} \times 100: \frac{1}{15} \times 100: \frac{8}{15} \times 100\)

40:6.67:53.3

⇒ \(\frac{40}{12}: \frac{6.67}{1}: \frac{53.3}{16}\)

⇒ \(\text { 1:2: } 1 \text { ie, } \quad \mathrm{CH}_2 \mathrm{O}\)

The percentage composition of an organic compound is as follows

C = 10.06, H = 0.84, Cl = 89.10

Question 29. Which of the following corresponds to the molecular formula if the vapor density is 60.0?

1. CH₃Cl
2. CHCl₃
3. CH₂Cl₂
4. None of these

Answer: 2. CHCl₃

Solution:

Molecular formula =(CHCl₃)1=CHCl₃

Question 30. If a compound on analysis was found to contain C=18.5%,H=1.55%,Cl=55.04% and O=24.81% then its empirical formula is

1. CH₂OCl
2. CH₂ClO₂
3. ClCH₂O
4. CHClO

Answer: 4. CHClO

Solution:

⇒ \(C: H: C l: O=\frac{18.5}{12}: \frac{1.55}{1}: \frac{55.04}{35.5}: \frac{24.81}{16}\)

= 1: 1: 1: 1

Question 31. 1.2g of organic compound of Kjeldahlization liberates ammonia which consumes 30 cm³ of 1N HCl. The percentage of nitrogen in the organic compound is

1. 30
2. 35
3. 46.67
4. 20.8

Answer: 2. 35

Solution:

Percentage of N in an organic compound = \(\frac{1.4 \times N \times V}{w}\)

⇒ \(=\frac{1.4 \times 1 \times 30}{1.2}=35\)

Question 32. The molecular mass of volatile substances may be obtained by

1. Bilstein method
2. Lassaigne’s method
3. Victor Mayer’s method
4. Liebig’s method

Answer: 3. Victor Mayer’s method

Solution: Victor Mayer’s method is applicable only for the determination of the molecular mass of a volatile substance

Question 33. 5.6 g of an organic compound on burning with excess of oxygen gave 17.6g of CO₂ and 7.2 g of H₂O. The organic compound is

1. C₆H₆
2. C₄H₈
3. C₃H₈
4. CH₃COOH

Answer: 2. C₄H₈

Solution:

Organic compound \(\begin{aligned}
& {[\mathrm{O}] \rightarrow \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}} \\
& 17.6 \mathrm{~g} \quad 7.2 \mathrm{~g}
\end{aligned}\)

⇒ \(\% \text { of } C=\frac{12}{44} \times \frac{17.6}{5.6} \times 100=85.7 \%\)

⇒ \(\% \text { of } H=\frac{2}{18} \times \frac{7.2}{5.6} \times 100=14.28 \%\)

Hence, the empirical formula of a compound of = CH₂

∴ Molecular formula of compound =C₄H₈

Question 34. 4 g of hydrocarbon on complete combustion gave 12.571 g of CO₂ and 5.143 g of water. What is its empirical formula?

1. CH
2. C₂H₃
3. CH₂
4. CH₃

Answer: 3. CH₂

Solution:

⇒ % C = \(\frac{12}{44} \times \frac{12.517}{4.0} \times 100=85.7\)

⇒ % H = \(\frac{2}{18} \times \frac{5.143}{4.0} \times 100=14.3\)

The mole ratio of C to H is \(\frac{85.7}{12}: \frac{14.3}{1}\)

⇒  \(7.14: 14.3=1: 2=\mathrm{CH}_2\)

Question 35. Liebig’s test is used to estimate

1. H
2. C
3. Both C and H
4. N

Answer: 3. Both C and H

Solution: Liebig’s method is used to estimate carbon and hydrogen. C and H [O]→ CO₂+H₂O

⇒ \(\% C=\frac{12}{44} \times \frac{\text { weight of } \mathrm{CO}_2}{\text { weight of compound }} \times 100\)

⇒ \(\% H=\frac{2}{18} \times \frac{\text { weight of } \mathrm{H}_2 \mathrm{O}}{\text { weight of compound }} \times 100\)

Question 36. An organic compound having a molecular mass of 60 was found to contain C=20%, H=6.67%, and N=46.67% while the rest is oxygen. On heating, it gave NH₃ along with a solid residue. The solid residue gave a violet color with an alkaline copper sulfate solution. The compound is

1. CH₃CH₂CONH₂
2. (NH₂)₂CO
3. CH₃CONH₂
4. CH₃NCO

Answer: 2. (NH₂)₂CO

Solution:

∴ Molecular formula=CH₄N₂O

Given compound gives a biuret test. Thus, the given compound is urea (NH₂)₂CO.

Assertion – Reasoning Type:

Each question contains Statement 1(Assertion) and Statement 2(Reason). Each question has 4 choices (1), (2), (3), and (4) out of which Only One is correct.

1. Statement 1 is True; Statement 2 is True; Statement 2 is the correct explanation for Statement 1
2. Statement 1 is True; Statement 2 is True; Statement 2 is not the correct explanation for Statement 1
3. Statement 1 is True, Statement 2 is False
4. Statement 1 is False, Statement 2 is True

Question 37.

1. Statement 1: ‘A Victor Mayer tube’ is of hard glass, having a side-tube, setup leading to the arrangement for the collection of displaced air over water.
2. Statement 2: An outer jacket of copper, containing a liquid boiling at nearly 30 higher than the substance whose molecular mass is to be determined

Answer: 2. Statement 1 is True; Statement 2 is True; Statement 2 is not the correct explanation for Statement 1

Solution: In Victor Mayer’s method, a known mass of the substance is converted into vapor by dropping it into a hot tube. The vapor displaces its own volume of air which is collected over water and its volume is measured at the observed temperature and pressure

Question 38.

1. Statement 1: Equivalent of K₂Cr₂O₇ has 1 equivalent of K and Cr and O each.
2. Statement 2: A species contains the same number of equivalents of its components.

Answer: 1. Statement 1 is True; Statement 2 is True; Statement 2 is the correct explanation for Statement 1

Solution: Fact.

Question 39.

1. Statement 1: Normality and molarity can be calculated from each other.
2. Statement 2: Normality is equal to the product of molarity and n.

Answer: 1. Statement 1 is True; Statement 2 is True; Statement 2 is the correct explanation for Statement 1

Solution: Normality =molarity× n

(n =mol wt. of solute /eq. wt. of solute)

Question 40.

Statement 1: In Messenger’s method, the colorless liquid is transferred to a beaker, and barium chloride is added to estimate the amount of sulphuric acid as BaSO₄ in the usual way

Statement 2: In an experiment, 0.36 g of an organic compound gave 0.35 of BaSO₄. the percentage of sulphur in the compound is 13.35%

Answer: 2. Statement 1 is True; Statement 2 is True; Statement 2 is not the correct explanation for Statement 1

Solution: Percentage of sulfur

⇒ \(=\frac{32}{233} \times \frac{0.35}{0.36} \times 100=13.35 \%\)

Question 41.

1. Statement 1: The molality of the solution does not change with a change in temperature.
2. Statement 2: The molality of the solution is expressed in units of moles per 1000 g of solvent.

Answer: 1. Statement 1 is True; Statement 2 is True; Statement 2 is the correct explanation for Statement 1

Solution: Molality does not depend upon the volume of the solution as molarity or normality. So, it does not depend upon temperature.

Question 42.

1. Statement 1: Equivalent weight of ozone in the change O₃→ O₂ is 8.
2. Statement 2: 1 mole O₃ of on decomposition gives 3/2 moles of O₂.

Answer: 2. Statement 1 is True; Statement 2 is True; Statement 2 is not the correct explanation for Statement 1

Solution:

⇒ \(\begin{array}{r}
2 \mathrm{O}_3 \rightarrow 3 \mathrm{O}_2 \\
2 \mathrm{~mol} \mathrm{O}_3 \equiv 3 \mathrm{~mol} \mathrm{O}_2=3 \times 2 \mathrm{eq} \mathrm{O} 2 \\
E_O=\frac{M}{6}=\frac{48}{6}=8
\end{array}\)

Question 43.

1. Statement 1: A solution that contains one gram equivalent of solute per liter of solutions is known as a molar solution.
2. Statement 2: Normality = \(\frac{mol.wt.of solute} {eq.wt.of solute}\)

Answer: 4. Statement 1 is False, Statement 2 is True

Solution: A solution; that contains one gram mole of solute per liter of solution is known as a molar solution (M).