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3 years ago

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The boiling point of an aqueous solution is 101.28 °c. what is the freezing point?

1 answer:

marissa [1.9K]3 years ago

3 0

Answer : -4.65 °C

Explanation : On solving the sum w.r.t to water as the aqueous solvent in solution;
(101.28°C - 100°C) / (0.512 c/m) = 2.5 m,

Now, 2.5 m X (1.86 °C/m) = 4.65 °C which will be the change in temperature of the solution in °C;

To find the freezing point we have to get it subtracted by 0°C

0°C - 4.65°C = -4.65°C

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Predict the products of the following reaction and balance the equation: ___Na2CO3(aq) + ___HCl (aq) ->

zheka24 [161]

Answer:

RxN: Na₂CO₃ (aq) + 2HCl (aq) → 2NaCl (aq) + H₂CO₃ (aq)

NIE: n/a

Explanation:

Step 1: Predict (Double Replacement)

Na₂CO₃ (aq) + HCl (aq) → NaCl (aq) + H₂CO₃ (aq)

Step 2: Balance

Na₂CO₃ (aq) + 2HCl (aq) → 2NaCl (aq) + H₂CO₃ (aq)

Step 3: Write Total Ionic Equation

2Na⁺ (aq) + CO₃²⁻ (aq) + 2H⁺ (aq) + 2Cl⁻ (aq) → 2Na⁺ (aq) + 2Cl⁻ (aq) + 2H⁺ (aq) + CO₃²⁻ (aq)

Step 4: Remove Spectator Ions

- Na⁺

- CO₃²⁻

- H⁺

- Cl⁻

Step 5: Write Net Ionic Equation

None.

3 0

3 years ago

What is the Molar Mass of 2.4g of oxygen

Anastasy [175]

<span>We have a sample of a gas which is a compound of carbon and oxygen. Chemical analysis shows that it contains 2.4 g of carbon and 6.4 g of oxygen. What is its formula?</span>

3 0

4 years ago

Someone can help me please??

kramer

C to D

Explanation:

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3 years ago

Read 2 more answers

The theoretical yield of a reaction is 75.0 grams of product and the actual yield is 42.0g. what is the percent yield? 75.0 56.0

Ronch [10]

I believe the percent yield is 56% if going by the formula of mass of actual yield/theoretical yield x 100%/

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4 years ago

A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.

Alexxandr [17]

<u>Answer:</u> The empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

<u>Explanation:</u>

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.

For Carbon = $\frac{0.313}{0.078}=4.01\approx 4$

For Hydrogen = $\frac{0.316}{0.078}=4.05\approx 4$

For Oxygen = $\frac{0.078}{0.078}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

$n=\frac{130g/mol}{68g/mol}=1.9\approx 2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2$

Hence, the empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

4 0

3 years ago