Question

What is the molarity of a solution in which 25.3 grams of potassium bromide is dissolved in 150. mL of solution?

Answer 1

Answer:

~1.417M

Explanation:

Molarity=(number of moles of solute)/(litres of solution)

In this case, we need to find moles of potassium bromide.

Mass=25.3g

Molar mass= 119g/mol

moles=(mass/molar mass)

        =(25.3)/(119)

        =0.2126moles of potassium bromide

Molarity=(0.2126)/(150/1000)

       ~1.417M

Hope this helps:)

Answer 2

Answer:

$\boxed {\boxed {\sf molarity \approx 1.42 \ M \ KBr}}$

Explanation:

Molarity is a measure of concentration in moles per liter.

$molarity= \frac{moles \ of \ solute}{ liters \ of \ solution}}$

1. Find Formula for Compound

We have the compound potassium bromide. Potassium (K) has an oxidation state of +1 and bromine (Br) has -1. They bond in a 1:1 ratio, so the formula is KBr.

2. Convert Grams to Moles

We are given the amount of solute in grams, but we need moles. To convert, we use the molar mass. These values are found on the Periodic Table. They are the same as the atomic masses, but the units are grams per moles (g/mol) instead of atomic mass units (amu).

Look up the individual element's molar mass.

• Potassium: 39.098 g/mol
• Bromine: 79.90 g/mol

The formula of KBr contains no subscripts, so we can add the molar masses.

• KBr: 39.098+ 79.90 =118.99 g/mol

Use the molar mass as a ratio.

$\frac {118.998 \ g\ KBr}{ 1 \ mol \ KBr}$

We want to convert 25.3 grams, so we multiply by that value.

$25.3 \ g\ KBr*\frac {118.998 \ g\ KBr}{ 1 \ mol \ KBr}$

Flip the ratio so the units of grams of KBr cancel.

$25.3 \ g\ KBr*\frac{ 1 \ mol \ KBr}{118.998 \ g\ KBr}$

$25.3*\frac{ 1 \ mol \ KBr}{118.998} = 0.2126086153\ mol \ KBr$

3. Convert Milliliters to Liters

Molarity uses liters, so we must convert the 150 milliliters. 1 liter contains 1000 milliliters.

$\frac{1 \ L }{1000 \ ml}$

$150 \ mL *\frac{1 \ L }{1000 \ ml}$

$150 *\frac{1 \ L }{1000}= 0.150 \ L$

4. Calculate Molarity

Now we have the moles of solute and liters of solution, so we can find molarity.

$molarity= \frac{ moles \ of \ solute}{liters \ of \ solution}$

$molarity= \frac{0.2126086153 \ mol \ KBr}{ 0.150 \ L}$

$molarity = 1.417390768\ mol KBr/L$

The original measurements had 3 significant figures, so our answer must have the same. For the number we found, that is the hundredth place. The 7 in the thousandth place tells us to round the 1 to a 2.

$molarity \approx 1.42 \ mol KBr/L$

1 mole per liter is equal to 1 Molar (M), so we must convert the units.

$molarity \approx 1.42 \ M \ KBr$