Question

What is the concentration in %m/v of a 0.617 M aqueous solution of methanol (MM = 32.04 g/mol)?

Answer 1

Answer:

The correct answer is  1.977 % m/v ≅ 2% m/v

Explanation:

We have:

0.617 M = 0.617 moles methanol/ 1 L solution

We need:

%m/v= grams of methanol/100 mL solution

So, first we convert the moles of methanol to grams by using the MM (32.04 g/mol). Then, we multiply by 0,1 to convert the volume in liters to 100 mL by using the ratio: 100 mL= 0.1 L:

0.617 mol / 1 L x 32.04 g/mol 0.1 L/100 mL= 1.977 g/100 mL= %m/v

Answer 2

The concentration in $\%_{m/v}$ of a 0.617 M aqueous solution of methanol is 1.98%.

To find the $%_{m/v}$ concentration of methanol we need to use the following equation:

$\%_{m/v} = \frac{m_{s}}{V_{sol}} \times 100$   (1)

Where:

$m_{s}$: is the mass of methanol in grams

$V_{sol}$: is the volume of the solution in milliliters

The molar concentration of methanol is:

$C = 0.617 M = 0.617 \:\frac{mol}{L}$

From this concentration, we can find the mass of methanol

$m = n*MM$   (2)

Where:

n: is the number of moles = C*V

MM: is the molar mass = 32.04 g/mol

Then, the mass of methanol is (eq 2):

$m = n*MM = C*V*MM = 0.617 mol/L*1 L*32.04 g/mol = 19.77 g$

Knowing that 1 L = 1000 mL, the $%_{m/v}$ concentration is (eq 1):

$\%_{m/v} = \frac{m_{s}}{V_{sol}} \times 100 = \frac{19.77 g}{1000 mL} \times 100 = 1.98 \%$

Therefore, the concentration in $\%_{m/v}$ is 1.98%.

Find more here:

• https://brainly.com/question/24958554?referrer=searchResults

• https://brainly.com/question/1386691?referrer=searchResults

I hope it helps you!