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2 years ago

36g of KOH dissolved in 800mL of water. What is the molality of the solution?

Chemistry

1 answer:

Bad White [126]2 years ago

5 0

Answer:

0.80m of KOH

Explanation:

Molality is an unit of concentration defined as the ratio between moles of solute and kg of solvent.

In the problem, the solute is KOH and solvent is water.

Moles of 36g KOH -Molar mass: 56.1g/mol- are:

36g KOH × (1mol / 56.1g) = 0.642 moles of KOH

Now, as density of water is 1g/mL, mass of 800mL of water is:

800mL × (1g / mL) × (1kg / 1000g) = 0.800kg of water

Thus, molality is:

0.642moles of KOH / 0.800kg = 0.80m of KOH

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A 3.96x10^-24 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000-cm cuvet; a blank solution containi

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Actual question from source:-

A 3.96x10-4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000 cm cuvette. A blank had an absorbance of 0.029. The absorbance of an unknown solution of compound A was 0.375. Find the concentration of A in the unknown.

Answer:

Molar absorptivity of compound A = $1502.53\ {Ms}^{-1}$

Explanation:

According to the Lambert's Beer law:-

$A=\epsilon l c$

Where, A is the absorbance

l is the path length

$\epsilon$ is the molar absorptivity

c is the concentration.

Given that:-

c = $3.96\times 10^{-4}\ M$

Path length = 1.000 cm

Absorbance observed = 0.624

Absorbance blank = 0.029

A = 0.624 - 0.029 = 0.595

So, applying the values in the Lambert Beer's law as shown below:-

$0.595=\epsilon\times 1.000\ cm\times 3.96\times 10^{-4}\ M$

$\epsilon=\frac{0.595}{3.96\times 10^{-4}}\ {Ms}^{-1}=1502.53\ {Ms}^{-1}$

Molar absorptivity of compound A = $1502.53\ {Ms}^{-1}$

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2 years ago

If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T

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Answer: The actual yield of $B_2O_3$ is 60.0 g

Explanation:-

The balanced chemical reaction :

$B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)$

Mass of $B_2H_6$ = $Density\times Volume=1.131g/ml\times 36.9ml=41.7g$

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles$

According to stoichiometry:

1 mole of $B_2H_6$ gives = 1 mole of $B_2O_3$

1.51 moles of $B_2H_6$ gives = $\frac{1}{1}\times 1.51=1.51$ moles of $B_2O_3$

Theoretical yield of $B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g$

Percent yield of $B_2O_3$ = $75.7\%$

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

$75.7\%=\frac{\text{Actual yield}}{79.3}\times 100$

${\text{Actual yield}}=60.0g$

Thus the actual yield of $B_2O_3$ is 60.0 g

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