What is the molarity of 20.2 g of potassium nitrate, KNO3, in enough water to make 250.0 mL of solution

A 7.27-gram sample of a compound is dissolved in 250. grams of benzene. The freezing point of this solution is 1.02°C below that

UkoKoshka [18]

Answer:

The correct answer is 146 g/mol

Explanation:

<em>Freezing point depression</em> is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:

ΔTf = Kf x m

Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:

ΔTf = 1.02ºC

Kf = 5.12ºC/m

From this, we can calculate the molality:

m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m

The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:

0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute

There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:

molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol

<em>Therefore, the molar mass of the compound is 146 g/mol </em>

6 0

3 years ago

Severus Snape explains that if you have the molar mass of a compound and the empirical formula of a compound you can determine t

KengaRu [80]

Answer:

The molecular formula of the compound : $C_4H_6O_2$

Explanation:

The empirical formula of the compound = $C_2H_3O$

The molecular formula of the compound = $C_{2n}H_{3n}O_n$

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 86 g/mol

Mass of empirical formula = 43 g/mol

Putting values in above equation, we get:

$n=\frac{86g/mol}{43g/mol}=2$

The molecular formula of the compound :

$C_{2\times 2}H_{3\times 2}O_2=C_4H_6O_2$

7 0

3 years ago

2. How many grams of NaCl are required to prepare 0.40 L of a 0.75 M solution?

Yuri [45]

The mass of a NaCl solution that is required to prepare 0.40 L of a 0.75 M solution is 17.55g. Details about mass can be found below.

<h3>How to calculate mass?</h3>

The mass of a substance can be calculated by multiplying the number of moles by its molar mass.

However, the number of moles of a solution must be initially calculated by using the following formula:

molarity = no of moles ÷ volume

no of moles = 0.75 × 0.40

no of moles = 0.3 moles

mass of NaCl = 0.3 × 58.5 = 17.55g

Therefore, the mass of a NaCl solution that is required to prepare 0.40 L of a 0.75 M solution is 17.55g.

Learn more about mass at: brainly.com/question/19694949

#SPJ1

4 0

2 years ago

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.76

aliina [53]

The given question is incomplete, the complete question is:

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?

Answer:

The lowest whole-number mass ratio in the two compounds is 1:2.

Explanation:

There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.

The no. of moles can be determined by using the formula,

moles = mass/molecular mass

moles = 2.98 g/207.2 g/mol

= 0.0144 moles

The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.

moles = 0.461 g /16 g/mol

= 0.0288 moles

The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2

On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,

moles = 9.89 g / 207.2 g/mol

= 0.0477 moles

The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,

moles = 0.763 g / 16 g

= 0.0477 moles

The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1

Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.

4 0

3 years ago

26. Balance the following equations:<br> Ca(s) + H3PO4(aq)Ca3(PO4)2(s) + H2(g)

yKpoI14uk [10]

Hey there!

Ca + H₃PO₄ → Ca₃(PO₄)₂ + H₂

Balance PO₄.

1 on the left, 2 on the right. Add a coefficient of 2 in front of H₃PO₄.

Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + H₂

Balance H.

6 on the left, 2 on the right. Add a coefficient of 3 in front of H₂.

Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂

Balance Ca.

1 on the right, 3 on the right. Add a coefficient of 3 in front of Ca.

3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂

Our final balanced equation:

3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂

Hope this helps!

3 0

3 years ago