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Yanka [14]
2 years ago
5

Calculate the molality of a solution that contains 51.2 g of naphthalene, C10H8, in 500 mL of carbon tetrachloride. The density

of CCl4 is 1.60 g/mL.
Chemistry
1 answer:
PtichkaEL [24]2 years ago
4 0

<u>Answer:</u> The molality of naphthalene solution is 0.499 m

<u>Explanation:</u>

Density is defined as the ratio of mass and volume of a substance.

\text{Density}=\frac{\text{Mass}}{\text{Volume}} ......(1)

Given values:

Volume of carbon tetrachloride = 500 mL

Density of carbon tetrachloride = 1.60 g/mL

Putting values in equation 1, we get:

\text{Mass of carbon tetrachloride}=(1.60g/mL\times 500mL)=800g

Molality is defined as the amount of solute expressed in the number of moles present per kilogram of solvent. The units of molarity are mol/kg. The formula used to calculate molarity:

\text{Molality of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent (in g)}} .....(2)

Given values:

Given mass of naphthalene = 51.2 g

Molar mass of naphthalene = 128.17 g/mol

Mass of solvent = 800 g

Putting values in equation 2, we get:

\text{Molality of naphthalene}=\frac{51.2\times 1000}{128.17\times 800}\\\\\text{Molality of naphthalene}=0.499m

Hence, the molality of naphthalene solution is 0.499 m

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Chloride ion is formed when chlorine atom gain one more electron. So, the ground-state electron configuration for the chloride ion is:-

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A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon
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<u>Answer:</u> The enthalpy of the reaction is 269.4 kJ/mol

<u>Explanation:</u>

To calculate the heat absorbed by the calorimeter, we use the equation:

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q = heat absorbed

c = heat capacity of calorimeter = 675 J/°C

\Delta T = change in temperature = T_2-T_1=(53.88-24.26)^oC=29.62^oC

Putting values in above equation, we get:

q_1=675J/^oC\times 29.62^oC=19993.5J

To calculate the heat absorbed by water, we use the equation:

q_2=mc\Delta T

where,

q = heat absorbed

m = mass of water = 925 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(53.88-24.26)^oC=29.62^oC

Putting values in above equation, we get:

q_2=925g\times 4.186J/g^oC\times 29.62^oC=114690.12J

Total heat absorbed = q_1+q_2

Total heat absorbed = [19993.5+114690.12]J=134683.62J=134.7kJ

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat absorbed = 134.7 kJ

n = number of moles of hydrocarbon = 0.500 moles

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{134.7kJ}{0.500mol}=269.4kJ/mol

Hence, the enthalpy of the reaction is 269.4 kJ/mol

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3 years ago
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