# What is the molarity of h2 so4 solution that has a density of 1.84 g cc and contains 98% by mass of h2 so4 given atomic mass of s 32?

The molar concentration unit [mol/ L (M)] is a conventionally widely used as concentration method. It is the number of moles of target substance (solute) dissolved in 1 liter of solution. Here is how to calculate the concentration.

(Weight of 1 liter solution) x (purity) ÷ molecular weight
[Specific gravity of solution (g/mL) x 1,000 (mL) x Purity (w/w%) /100 ÷ Molecular weight]

For example, let's calculate the molar concentration of 2-mercaptoethanol (HSCH2CH2OH). The necessary information is as follows.

• Specific gravity (or density) = 1.114 g/mL
• Purity (or content) = 100 w/w% (assumed)
• Molecular weight = 78.13

By calculating this value by applying this value to the above equation, you can know the molar concentration.
1.114 g/mL x 1,000mL x 100w/w%/100 ÷ 78.13 = 14.26mol/L

In order to caluculate the concentration like above, it is necessary to know three points of "specific gravity (or density)", "purity (or content)" and "molecular weight". The table below is a quick reference chart of common acid and base concentrations. In acid and alkali, there is a use for "neutralization titration", "normality (N)" is often used.

【Quick reference chart of common acid and base concentrations】

Compound Molecular formula Molecular weight Purity
(w/w％)
Specific gravity
(20℃)
Concentration
(mol/L)
Equivalent Normality
(N)
Hydrochloric acid HCl 36.46 20% 1.10 6.0 1 6.0
35% 1.17 11.2 11.2
Nitric acid HNO3 63.01 60% 1.37 13.0 1 13.0
65% 1.39 14.3 14.3
70% 1.41 15.7 15.7
Sulfate H2SO4 98.08 100% 1.83 18.7 2 37.3
Phosphoric acid H3PO4 98.00 85% 1.69 14.7 3 44.0
90% 1.75 16.1 48.2
Acetate CH3COOH 60.05 100% 1.05 17.5 1 17.5
Perchloric acid HClO4 100.46 60% 1.54 9.2 1 9.2
70% 1.67 11.6 11.6
Hydrogen peroxide water H2O2 34.01 30% 1.11 9.8 -
35% 1.13 11.6
Ammonia water NH3 17.03 25% 0.91 13.4 1 13.4
28% 0.90 14.8 14.8

【Quick reference of concentration and unit】

●How to express concentration of solution

Expression Commentary
Weight percent concentration "g number" of solute in 100g solution. Expressed as w/w%, wt%, and % for density in many cases.
Volume percent concentration "m number" of solute in 100m solution. Expressed as v/v% when mixture or solute is liquid.
Weight versus volume percent concentration "g number of solute in 100m of solution. Expressed as w/v%.
Normality Gram equivalent number of solute in 1L solution. Expressed as N for capacity analysis.
Volume specific concentration Concentration indirectly expressed by the volume ratio of diluting the liquid reagent. It is used in JIS and others.
Example: Sulfuric acid (1 + 2) → Sulfuric acid is shown diluted with 2 volumes of water.
Weight ratio concentration Concentration indirectly expressed by weight ratio at which solid reagent is dissolved. It is used in JIS and others.
Example: Sodium chloride (1 + 19) →Dissolved in 19 weight of water with respect to 1 of NaCl.
Molarity Mol number of target substance (solute) in 1L of solution. Expressed as mol/ or M.

●Prefix representing multiple

Express bigness Express smallness
100 =102 h（Hecto） 1/100 =10-2 c（Centi） ％（Percent）
1000 =103 k（Kilo） 1/1000 =10-3 m（Milli） ‰（Permili）
100万 =106 M（Mega） 1/100万 =10-6 μ（Micro） ppm
1 Billion =109 G（Giga） 1/10Billion =10-9 n（Nano） ppb
1 Trillion =1012 T（Tera） 1/1 Trillion =10-12 p（Pico） ppt
1000 Trillion =1015 p（Peta） 1/1000 Trillion =10-15 f（Femto） ppq

●ppmConversion table

ppb ppm % mg/g mg/L
1,000 1 0.0001 0.001 1
10,000 10 0.001 0.01 10
100,000 100 0.01 0.1 100
1,000,000 1,000 0.1 1 1,000
10,000,000 10,000 1 10 10,000

Verified

Hint :In the mole concept we have studied about the molarity and we have studied that molar concentration of substance is also known as molarity. It is defined as the ratio of moles of substance to volume of solution in litre.

> Mole is defined as weight in grams divided by the molecular weight of the substance. So we will find out the molarity by the formula; M=Number of moles of solute /volume of the solution in litre.> As it is given in the problem that ${H_2}S{O_4}$ is $98\%$ by weight that means that $100g$ solution contains $98g$${H_2}S{O_4} by mass.> Density of solution of sulphuric acid is given that is 1.84g/cc; Density = \dfrac{{mass}}{{Volume}} \Rightarrow Volume = \dfrac{{mass}}{{Density}}$$ \Rightarrow Volume = \dfrac{{100}}{{1.84 \times 1000}}L$$\Rightarrow Volume = 0.0543 Litre> Molecular weight of sulphuric acid ({H_2}S{O_4}) = 2×atomic weight of hydrogen atomic weight of sulphur +4×atomic weight of oxygenMolecular weight of sulphuric acid ({H_2}S{O_4}) = 2 \times 1 + 32 + 4 \times 16$$ = 98$- Now here weight of sulphuric acid in gram is $98g$ and molecular weight of sulphuric acid is $98g/mol$ - So first calculate the number of moles before calculating molarity. Number of moles of ${H_2}S{O_4}$=Weight in gram of ${H_2}S{O_4}$/ molecular weight of ${H_2}S{O_4}$$= 98/98$$ = 1$- Hence the molarity of ${H_2}S{O_4}$=Number of moles /Volume in litre $=\dfrac{1}{{0.0543}}$$= 18.4M$So the correct option is (c) that is $18.4M$.Note : In mole problems we should always remember some important relation .Here in this problem we have used some relations to solve the problem as we have found the mass with the help of given density and then put the value of mass in the formula to find the number of moles. Once the number of moles is determined we calculate molarity of sulphuric acid easily with the help of molarity formula.

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What is the molarity of H 2 SO 4 solution that has a density of 1.84 g / cc and contains 98 % by mass of H 2 SO 4

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What is the molarity of H 2 SO 4 solution that has a density of 1.84 g / cc and contains 98 % by mass of H 2 SO 4

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