Question 4 of 10 The Fourteenth Amendment says, "No State shall ... deny to any person
within its jurisdiction the equal protection of the laws. 2
How did the Supreme Court expand this protection in the 1950s?
A. It interpreted laws to include religious beliefs.
B. It said that state jurisdictions included adjoining states.
C. It ruled that corporations qualified as people.
D. It applied the protection to the federal government as well as to
the states.
SUENIT

\boxed {\boxed {\sf molarity \approx 1.42 \ M \ KBr}}molarity≈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}1 mol KBr118.998 g 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}25.3 g KBr∗1 mol KBr118.998 g 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 g KBr∗118.998 g KBr1 mol KBr

25.3*\frac{ 1 \ mol \ KBr}{118.998} = 0.2126086153\ mol \ KBr25.3∗118.9981 mol KBr=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}1000 ml1 L

150 \ mL *\frac{1 \ L }{1000 \ ml}150 mL∗1000 ml1 L

150 *\frac{1 \ L }{1000}= 0.150 \ L150∗10001 L=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=liters of solutionmoles of solute

molarity= \frac{0.2126086153 \ mol \ KBr}{ 0.150 \ L}molarity=0.150 L0.2126086153 mol KBr

molarity = 1.417390768\ mol KBr/Lmolarity=1.417390768 molKBr/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/Lmolarity≈1.42 molKBr/L

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

molarity \approx 1.42 \ M \ KBrmolarity≈1.42 M KBr

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