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4 years ago

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Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0

.771 g of biphenyl (C12H10) in 25.0 g of cyclohexane?

1 answer:

Elza [17]4 years ago

4 0

Answer: $2.49^0C$

Explanation:

Depression in freezing point is:

$T_f^0-T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of solvent (cyclohexane) = $6.50^oC$

$k_f$ = freezing point constant of solvent (cyclohexane) = $20.0^oC/m$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute (biphenyl) = 0.771 g

$w_1$ = mass of solvent (cyclohexane) = 25.0 g

$M_2$ = molar mass of solute (biphenyl) =

Now put all the given values in the above formula, we get:

$(6.50-T_f)^oC=1\times (20.0^oC/m)\times \frac{(0.771g)\times 1000}{154\times (25.0g)}$

$(6.50-T_f)^oC=4.01$

$T_f=2.49^0C$

Therefore, the freezing point of a solution made by dissolving 0.771 g of biphenyl in 25.0 g of cyclohexane is $2.49^0C$

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How many moles are there in 44.71 g of chlorine (cl)? use significant figures?

krok68 [10]

Mass of chlorine = $44.71 g$ (given)

Molar mass of chlorine = $35.45 g/mol$

Now, number of moles is defined as the ratio of given mass in gram to the molar mass of the compound.

Number of moles = $\frac{given mass in g}{molar mass}$

Number of moles of chlorine = $\frac{44.71 g}{35.45 g/mol}$

= $1.26121297 moles$

Now, according to significant figure rules, the least number of significant figure in the question identifies the number of significant figures in the solution.

Thus, in question, both digit have four significant figures, therefore, the answer must be in four significant figures.

Hence, number of moles of chlorine = $1.261 moles$

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4 years ago

Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting

larisa [96]

Answer: Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

Explanation:

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $2.3\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$2.3\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.28\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.28\times 10^{-5}M$

2) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.28\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.28\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.034\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.034\times 10^{-5}M$

3) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.034\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.034\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=4.0\times 10^{-8}M$

Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

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almond37 [142]

Answer:

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Solubility is dependent on temperature. This is the reason why we must state the particular temperature at which a given solubility data was collected. Above or below the quoted temperature, the solubility data may no longer be accurate.

Let's return to the question, at elevated temperatures, we can see that the KCl is very soluble in water as evidenced by the clear solution obtained at high temperature. However, as the temperature decreases, the solubility of KCl also decreases accordingly and recrystallization of the solute occurred hence the formation of a cloudy solution and the settling of some solid at the bottom of the flask.

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Thy answer tis
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ahh but why you may ask?
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the simplest answer,
because chemistry.

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