Answer:
3g of NaI is needed to prepare 400.0 mL of 0.0500 M NaI solution.
Explanation:
A 0.0500M solution means that is 0.0500 moles of NaI per every 1000 mL of solution. Thus, to prepare 400mL:
[tex]\frac{400mL}{1000mL}x0.0500moles=0.0200moles[/tex]
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:
[tex]150\frac{g}{mol}x0.0200moles = 3g[/tex]
Answer:
We need 3.00 grams NaI
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
