All categories

2 years ago

What is the molarity of a solution in which 7.1 g of sodium sulfate is dissolved in enough water to make 100. mL of solution?

Chemistry

2 answers:

Mamont248 [21]2 years ago

6 0

Answer:

0.5 M

Explanation:

Hello,

In this case, since sodium sulfate molar mass is 142.04 g/mol we first need to compute the moles in 7.1 g as shown below:

$n=7.1g*\frac{1mol}{142.04 g} =0.050mol$

Then, since 100 mL are 0.1 L the molarity turns out:

$M=\frac{n}{V}=\frac{0.05mol}{0.1L}\\\\M=0.5M$

Regards.

Sidana [21]2 years ago

3 0

Answer:

0.50 M

Explanation:

Given data

Mass of sodium sulfate (solute): 7.1 g

Volume of solution: 100 mL

Step 1: Calculate the moles of the solute

The molar mass of sodium sulfate is 142.04 g/mol. The moles corresponding to 7.1 grams of sodium sulfate are:

$7.1g \times \frac{1mol}{142.02g} = 0.050mol$

Step 2: Convert the volume of solution to liters

We will use the relation 1 L = 1000 mL.

$100mL \times \frac{1L}{1000mL} =0.100L$

Step 3: Calculate the molarity of the solution

$M = \frac{moles\ of\ solute }{liters\ of\ solution} = \frac{0.050mol}{0.100L} =0.50 M$

You might be interested in

Which of the following is the ground state electrons configuration of the F ion

Vinil7 [7]

Answer:

That leaves us with: 1s2 2s2 2p6 And that is the electron configuration for F- , the Fluoride ion.

Explanation:

plz give brainliest

3 0

2 years ago

Why is acylation reagent important.

shepuryov [24]

Answer:

Acylation can be used to prevent rearrangement reactions that would normally occur in alkylation.

3 0

2 years ago

Could I have help on 4 and 5? Just the letter please I don't need an explaintion

PIT_PIT [208]

5. is b
but im not sure about 4
hope this helps

6 0

3 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

Answer: The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

Explanation:

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

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

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago

Which student is responses are correct?

irga5000 [103]

Answer:

Neutrons: Electron cloud

Explanation:

7 0

3 years ago