Question

C. 49.8 grams of ki is dissolved in enough water to make 1.00 l of solution. what is the molarity?

Answer 1

Hey There!:

Molar Mass KI => 166.003 g/mol

* number of moles:

n = mass of solute / molar mass

n = 49.8 / 166.003

n = 0.3 moles KI

Therefore:

M = n / V

M = 0.3 / 1.00

M = 0.3 mol/L

hope this helps!

Answer 2

Answer:

49.8 grams of KI is dissolved in 1 L of solution have a molarity = 0.3M

Explanation:

We need the molar mass of KI, we can find the molar mass with the atomic weight of potassium and iodo. We can find this information using the periodic table of the elements.

Atomic weight of $K = 39.0983\frac{g}{mol}$

Atomic weight of $I = 126.90447\frac{g}{mol}$

Adding the atomic weights of potassium and iodo, we can obtain the molar mass of KI:

$KI= 39.0983\frac{g}{mol} +126.90447\frac{g}{mol} \\ KI= 166.003\frac{g}{mol}$

To find the molarity we need the moles of KI in 49.8 g of KI

$moles_K_I=49.8g*\frac{1mol}{166.003 g} \\moles_K_I= 0.3$

The molarity formula is:

$M=\frac{n}{V}$

Where:

n= moles of KI

V= volume

$M=\frac{0.3mol}{1L} = 0.3M$

The molarity is 0.3 M