Question

. How many grams of CaCl₂ (molar mass = 111

g/mol) are needed to prepare 100. mL of 0.100
M solution? Show your calculation to receive
full credit.
A) 55.5 g
B) 111 g
C) 222 g
D) 555 g

Answer 1

1.11 grams CaCl₂

**I am aware that this is not one of the answer choices, but I pretty confident that this should be the correct answer according the the given values

Step-by-step explanation:

To find the mass of CaCl₂ necessary, you need to

(1) convert the volume from mL to L (1,000 mL = 1 L)

(2) calculate the amount of moles (M = moles / L)

(3) convert moles to grams (using the molar mass)

It is important to arrange the conversions/ratios in a way that allows for the cancellation of units. The final answer should have 3 sig figs like the given values.

(Step 1)

100. mL CaCl₂ 1 L
-------------------------- x ------------------- = 0.100 L CaCl₂
1,000 mL

(Step 2)

Molarity (M) = moles / volume (L) <----- Molarity ratio

0.100 M = moles / 0.100 L <----- Insert values

0.0100 = moles <----- Multiply both sides by 0.100 L

(Step 3)

Molar Mass (CaCl₂): 111 g/mol

0.0100 moles CaCl₂ 111 grams
--------------------------------- x --------------------- = 1.11 grams CaCl₂
1 mole