Question

How many grams of NaI are needed to prepare 400.0 mL of 0.0500 M NaI solution?

Answer 1

We need 3.00 grams NaI

Step-by-step explanation:

Step 1: Data given

Volume = 400.0 mL = 0.400L

Molarity = 0.0500 M

Molar mass of NaI = 149.89 g/mol

Step 2: Calculate moles

Moles NaI = molarity * volume

Moles NaI = 0.0500 M * 0.400 L

Moles NaI = 0.02 moles

Step 3: Calculate mass NaI

Mass NaI = moles NaI * molar mass NaI

Mass NaI = 0.02 moles * 149.89 g/mol

Mass NaI = 3.00 grams

We need 3.00 grams NaI

Answer 2

3g of NaI is needed to prepare 400.0 mL of 0.0500 M NaI solution.

Step-by-step explanation:

A 0.0500M solution means that is 0.0500 moles of NaI per every 1000 mL of solution. Thus, to prepare 400mL:

`(400mL)/(1000mL)x0.0500moles=0.0200moles`

Also, the molar mass of NaI is 127g/mol + 23g/mol = 150g/mol

Consequently, multiplying the molar mass of NaI by the moles the necessary mass is obtained:

`150(g)/(mol)x0.0200moles = 3g`