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3 years ago

What is the molarity of a solution composed of 5.85 g of potassium iodide, KI, dissolved

Chemistry

1 answer:

Troyanec [42]3 years ago

7 0

Answer:

0.282 M

General Formulas and Concepts:

Chemistry - Solutions

Reading a Periodic Table
Using Dimensional Analysis
Molarity = moles of solute / liters of solution

Explanation:

Step 1: Define

5.85 g KI

0.125 L

Step 2: Identify Conversions

Molar Mass of K - 39.10 g/mol

Molar Mass of I - 126.90 g/mol

Molar Mass of KI - 39.10 + 126.90 = 166 g/mol

Step 3: Convert

$5.85 \ g \ KI(\frac{1 \ mol \ KI}{166 \ g \ KI} )$ = 0.035241 mol KI

Step 4: Find Molarity

M = 0.035241 mol KI / 0.125 L

M = 0.281928

Step 5: Check

We are given 3 sig figs. Follow sig fig rules and round.

0.281928 M ≈ 0.282 M

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How many moles of potassium are contained in 150 g of potassium

BaLLatris [955]

Answer:

150 g of potassium contained 3.8 moles of potassium.

Explanation:

Given data:

Mass of potassium = 150 g

Moles of potassium = ?

Solution:

Number of moles = mass/ molar mass

Molar mass of potassium = 39 g/mol

Now we will put the values in formula:

Number of moles = mass/ molar mass

Number of moles = 150 g/ 39 g/mol

Number of moles = 3.8 mol

150 g of potassium contained 3.8 moles of potassium.

3 0

4 years ago

If an atom has 7 protons, 8 neutrons and 9 electrons, what is the mass number?

Brums [2.3K]

Mass = protons + neutrons
7+8 = 15 g

3 0

3 years ago

Read 2 more answers

A student dissolves of 15 g aniline in of a solvent with a density of . The student notices that the volume of the solvent does

VikaD [51]

The given question is incomplete. The complete question is ;

A student dissolves of 15 g aniline in 200 ml of a solvent with a density of 1.05 g/ml. The student notices that the volume of the solvent does not change when the aniline dissolves in it. Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits.

Answer: The molarity is 0.81 M and molality is 0.82 m

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute

$V_s$ = volume of solution in ml = 200 ml

${\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{15g}{93g/mol}=0.16mol$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{0.16\times 1000}{200}=0.81M$

Thus molarity is 0.81 M

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molarity=\frac{n\times 1000}{W_s}$

where,

n = moles of solute

$W_s$ = weight of solvent in g

${\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{15g}{93g/mol}=0.16mol$

Mass of solution = $Density\times Volume=1.05g/ml\times 200ml=210g$

mass of solvent = mass of solution - mass of solute = (210 - 15) g = 195 g

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.16\times 1000}{195g}=0.82mole/kg$

Therefore, the molality of solution is 0.82m

3 0

3 years ago

What is the conjugate acid base pair for H20+HBR->H30+BR

True [87]

Answer:

TABLE OF CONJUGATE ACID-BASE PAIRS Acid BaseK a (25oC) HClO 4ClO 4 – H 2 SO 4HSO 4 – HCl Cl– HNO 3NO 3 – H 3 O +H 2 O H 2 CrO 4HCrO 4 –1.8 x 10–1

Explanation:

4 0

3 years ago

What weight of sodium formate must be added to 4.00L of 1.00 M formic acid to produce a buffer solution that has a pH of 3.50?

tester [92]

Answer:

156,4 g of sodium formate

Explanation:

The pka of the formic acid is 3,74. Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA] (1)

Where A⁻ is the conjugate base (Formate) and HA is the formic acid.

4.00L of 1.00M formic acid contain:

4.00L × (1.00mol /L) = 4.00 moles

Replacing these moles, the desired pH and the pka value in (1):

3,50 = 3,74 log₁₀ [A⁻] / 4,00 moles

-0,24 = log₁₀ [A⁻] / 4,00 moles

0,575 = [A⁻] / 4,00 moles

2,30 moles = [A⁻]

That means you need 2,30 moles of formate (Sodium formate), to produce the desired buffer. As the molar mass of sodium formate is 68,01g/mol, the weight of sodium formate that must be added is:

2,30mol×(68,01g/mol) = 156,4 g of sodium formate.

I hope it helps!

7 0

3 years ago