Ask question

Ask question

All categories

3 years ago

13

When 3.24 g of a nonelectrolyte solute is dissolved in water to make 955 mL of solution at 22 °C, the solution exerts an osmotic

pressure of 973 torr. What is the molar concentration of the solution? concentration: M How many moles of solute are in the solution? moles of solute: mol What is the molar mass of the solute? molar mass: g / mol

1 answer:

attashe74 [19]3 years ago

4 0

Answer:

The molar concentration is 0.0529 M

We have 0.0505 moles solute

The molar mass of the solute is 64.16 g/mol

Explanation:

Step 1: Data given

Mass of the nonelectrolyte solute = 3.24 grams

Volume of water = 955 mL

Temperature = 22 °C = 295 K

Osmotic pressure = 973 torr = 973 / 760 = 1.28026 atm

Step 2: Calculate mass water

Mass water = 955 mL * 1g/mL

Mass water = 955 grams = 0.955 kg

Step 3: Calculate the molar concentration

π = i*M*R*T

⇒with π = the osmotic pressure = 1.28026 atm

⇒with i = the van't Hoff factor = 1

⇒with M = the molar concentration

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 295 K

M = 1.28026 / (0.08206*295)

M = 0.0529 M

Step 4: Calculate moles solute

Molar concentration = moles / volume

Moles solute = Molar concentration * volume

Moles solute = 0.0529 M * 0.955 L

Moles solute = 0.0505 moles

Step 5: Calculate molar mass

Molar mass = mass / moles

Molar mass = 3.24 grams / 0.0505 moles

Molar mass = 64.16 g/mol

The molar concentration is 0.0529 M

We have 0.0505 moles solute

The molar mass of the solute is 64.16 g/mol

You might be interested in

How many molecules of carbon dioxide are dissolved in 0.550 L of water at 25 °C if the pressure of CO2 above the water is 0.250

Grace [21]

<u>Answer:</u> The number of molecules of carbon dioxide gas are $2.815\times 10^{21}$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{CO_2}=K_H\times p_{CO_2}$

where,

$K_H$ = Henry's constant = $0.034mol/L.atm$

$C_{CO_2}$ = molar solubility of carbon dioxide gas

$p_{CO_2}$ = pressure of carbon dioxide gas = 0.250 atm

Putting values in above equation, we get:

$C_{CO_2}=0.034mol/L.atm\times 0.250atm\\\\C_{CO_2}=8.5\times 10^{-3}M$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of carbon dioxide = $8.5\times 10^{-5}M$

Volume of solution = 0.550 L

Putting values in above equation, we get:

$8.5\times 10^{-3}M=\frac{\text{Moles of }CO_2}{0.550L}\\\\\text{Moles of }CO_2=(8.5\times 10^{-3}mol/L\times 0.550L)=4.675\times 10^{-3}mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules

So, $4.675\times 10^{-3}$ moles of carbon dioxide will contain = $(6.022\times 10^{23}\times 4.675\times 10^{-3})=2.815\times 10^{21}$ number of molecules

Hence, the number of molecules of carbon dioxide gas are $2.815\times 10^{21}$

3 0

3 years ago

Read 2 more answers

An unknown amount of acid can often be determined by adding an excess of base and then back-titrating the excess. A 0.3471−g sam

MAXImum [283]

Explanation:

The given data is as follows.

Mass of mixture = 0.3471 g

As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.

This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.

Hence, moles of NaOH in 97 ml = $\frac{0.1090 \times 97}{1000}$

= $10.573 \times 10^{-3} mol$

Moles of HCl in 21.00 ml = $\frac{0.2060 \times 21}{1000}$

= $4.326 \times 10^{-3}$ mol

Therefore, total moles of NaOH that reacted are as follows.

$10.573 \times 10^{-3} mol$ - $4.326 \times 10^{-3} mol$

= $6.247 \times 10^{-3}$ mol

So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.

$\frac{6.247 \times 10^{-3}}{3}$

= $2.082 \times 10^{-3}$ mol

Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.

$2.082 \times 10^{-3} mol \times 40 g/mol$

= $83.293 \times 10^{-3}$ g

= 0.0832 g

Whereas molar mass of benzoic acid is 122 g/mol.

Therefore, 40 g NaOH = 122 g benzoic acid

So, 0.0832 g NaOH = $\frac{122 g}{40 g} \times 0.0832 g$

= 0.253 g

Hence, calculate the % mass of benzoic acid as follows.

$\frac{0.253 g}{0.3471 g} \times 100$

= 73.10%

Thus, we can conclude that mass % of benzoic acid is 73.10%.

4 0

3 years ago

If 42 grams of carbon and 52 grams of oxygen are used, how many grams of CO2 will be produced (Hint: find the

zavuch27 [327]

Answer:

71.5g

Explanation:

The reaction equation is given as:

C + O₂ → CO₂

Mass of C = 42g

Mass of O₂ = 52g

Unknown:

Mass of CO₂ produced = ?

Solution

Now to solve this problem, we have to find limiting reactant which is the one given in short supply in this reaction.

The extent of the reaction is controlled by this reactant.

Find the number of moles of the given species;

Number of moles = $\frac{mass}{molar mass}$

Number of moles of C = $\frac{42}{12}$ = 3.5mol

Number of moles of O₂ = $\frac{52}{32}$ = 1.63mol

Now;

From the balanced reaction equation;

1 mole of C reacted with 1 mole of O₂

We see that C is in excess and O₂ is the limiting reactant.

1 mole of O₂ will produce 1 mole of CO₂

So; 1.63mole of O₂ will produce 1.63 mole of CO₂

Mass of CO₂ = number of moles x molar mass

Molar mass of CO₂ = 44g/mol

Mass of CO₂ = 1.63 x 44 = 71.5g

8 0

2 years ago

Is facilitated diffusion passive or active transport?

bazaltina [42]

Answer:

Its passive

Explanation:

Facilitated diffusion is a type of passive transport that allows substances to cross membranes with the assistance of special transport proteins.

6 0

2 years ago

In a chemical equation, a(n) _____ tells one how many atoms or molecules of a reactant or a product take part in a reaction.

aleksandrvk [35]

In a chemical equation, a subscript tells one how many atoms or molecules of a reactant or a product take part in a reaction.

8 0

3 years ago