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3 years ago

A 3.0 L solution contains 73.5 g of H2SO4. Calculate the molar concentration of the solution.

1 answer:

6 0

Molar mass :

$H_2SO_4= 1*2+32+16*4= 98g/mol$

Number of moles:

$n = \frac{solute}{molar( mass)}$

$n = \frac{73.5g}{98g/mol}$

$n=0.75 (moles)$

Volume = 3.0 L

$M = \frac{n}{V}$

$M = \frac{0.75(moles)}{3.0L}$

$M = 0.25 mol/L ^{-1}$

hope this helps!

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The equation used to calculate enthalpy change is of a reaction is:

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Putting values in above equation, we get:

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To calculate the number of moles, we use ideal gas equation, which is:

$PV=nRT$

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = $0.0831\text{ L. bar }mol^{-1}K^{-1}$

T = temperature of the mixture = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol$

To calculate the heat released of the reaction, we use the equation:

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

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

$-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ$

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