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Allushta [10]
3 years ago
6

A chemical supply company sells a concentrated solution of aqueous h2so4 (molar mass 98 g mol−1 ) that is 50. percent h2so4 by m

ass. at 25°c, the density of the solution is 1.4 g ml−1 . what is the molarity of the h2so4 solution at 25°c?
Chemistry
1 answer:
trapecia [35]3 years ago
8 0
Answer is: <span>the molarity of the sulfuric acid is 7.14 M.
</span>ω(H₂SO₄) = 50% ÷ 100% = 0.5.<span>
d(H</span>₂SO₄) = 1.4 g/mL.
V(H₂SO₄) = 100 mL ÷ 1000 mL/L = 0.1 L..
mr(H₂SO₄) = d(H₂SO₄) · V(H₂SO₄).
mr(H₂SO₄) = 1.4 g/mL · 100 mL.
mr(H₂SO₄) = 140 g.
m(H₂SO₄) = ω(H₂SO₄) · mr(H₂SO₄).
m(H₂SO₄) = 0.5 · 140 g.
m(H₂SO₄) = 70 g.
n(H₂SO₄) = m(H₂SO₄) ÷ M(H₂SO₄).
n(H₂SO₄) = 70 g ÷ 98 g/mol.
n(H₂SO₄) = 0.714 mol.
c(H₂SO₄) = n(H₂SO₄) ÷ V(H₂SO₄).
c(H₂SO₄) = 0.714 mol ÷ 0.1 L.
C(H₂SO₄) = 7.14 M.
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A 0.1014 g sample of a purified compound containing C, H, and, O was burned in a combustion apparatus and produced 0.1486 g CO2
Alina [70]

Answer: The empirical formula for the given compound is CH_2O

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=0.1486g

Mass of H_2O=0.0609g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 0.1486 g of carbon dioxide, \frac{12}{44}\times 0.1486=0.0405g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.0609 g of water, \frac{2}{18}\times 0.0609=0.00677 of hydrogen will be contained.

Mass of oxygen in the compound = (0.1014) - (0.0405 + 0.00677) = 0.054 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.0405g}{12g/mole}=0.003375moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.00677g}{1g/mole}=0.00677moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.054g}{16g/mole}=0.003375moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = \frac{0.003375}{0.003375}=1

For Hydrogen  = \frac{0.00677}{0.003375}=2.00\times 2

For Oxygen  = \frac{0.003375}{0.003375}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 2 : 1

The empirical formula for the given compound is C_1H_2O_1=CH_2O

Thus, the empirical formula for the given compound is CH_2O

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