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2 years ago

Calculate the mass percent by volume of 330.1 g of glucose (C₆H₁₂O₆, MM = 180.2 g/mol) in 325 mL of solution.

Chemistry

1 answer:

nadezda [96]2 years ago

3 0

Answer: The mass percent by volume is 101.6%

Explanation:

The solution concentration expressed in percent by volume means that the amount of solute present in 100 parts volume of solution.

It is represented in formula as :

mass percent by volume = $\frac{\text {mass of solute}\times 100}{\text {Volume of solution in ml}}\%$

Given : mass of glucose = 330.1 g

volume of solution = 325 ml

Thus mass percent by volume = $\frac{330.1g\times 100}{325ml}=101.6\%$

Thus the mass percent by volume is 101.6%

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For some reaction carried out at constant atmospheric pressure and at a constant temperature of 25◦C, it is found that ∆H = −38.

kompoz [17]

Answer: ΔG =23.169kJ/mol

Explanation:

Solution

To Calculate Gibbs free energy ΔG for the reaction above we use the equation ΔG=ΔH−TΔS.

Where

ΔH= 38.468 kJ/mol = 38468 J/mol

∆S = +51.4 J mol−1 K−1).

T = 25◦C =298k

ΔG= 38468J/mol−298k(51.4 J mol−1 K−1).

ΔG = 38468 J/mol - 15317.2J/mol

ΔG = 23168.8J/mol

ΔG =23.169kJ/mol

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3 years ago

The ksp of manganese(ii) carbonate, mnco3, is 2.42 × 10-11. calculate the solubility of this compound in g/l.

Vikki [24]

Answer : The solubility of this compound in g/L is $565.414\times 10^{-6}g/L$ .

Solution : Given,

$K_{sp}=2.42\times 10^{-11}$

Molar mass of $MnCO_3$ = 114.945g/mole

The balanced equilibrium reaction is,

$MnCO_3\rightleftharpoons Mn^{2+}+CO^{2-}_3$

At equilibrium s s

The expression for solubility constant is,

$K_{sp}=[Mn^{2+}][CO^{2-}_3]$

Now put the given values in this expression, we get

$2.42\times 10^{-11}=(s)(s)\\2.42\times 10^{-11}=s^2\\s=0.4919\times 10^{-5}=4.919\times 10^{-6}moles/L$

The value of 's' is the molar concentration of manganese ion and carbonate ion.

Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.

$s=4.919\times 10^{-6}moles/L\times 114.945g/mole=565.414\times 10^{-6}g/L$

Therefore, the solubility of this compound in g/L is $565.414\times 10^{-6}g/L$ .

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