Answer:
Molarity = 5.22 M
Explanation:
Given data:
Mass of sodium chloride = 7.0 g
Volume of solution = 23.0 mL ( 23.0/1000 = 0.023 L)
Molarity = ?
Solution;
Number of moles of NaCl = 7.0 g/ 58.4 g/mol
Number of moles of NaCl = 0.12 mol
Molarity = moles of solute / volume in litter
Molarity = 0.12 mol / 0.023 L
Molarity = 5.22 M
Answer:
3.33 M
Explanation:
It seems your question is incomplete, however, that same fragment has been found somewhere else in the web:
" <em>A chemist prepares a solution of silver nitrate (AgNO3) by measuring out 85.g of silver nitrate into a 150.mL volumetric flask and filling the flask to the mark with water.</em>
<em>Calculate the concentration in mol/L of the chemist's silver nitrate solution. Be sure your answer has the correct number of significant digits.</em> "
In this case, first we <u>calculate the moles of AgNO₃</u>, using its molecular weight:
- 85.0 g AgNO₃ ÷ 169.87 g/mol = 0.500 mol AgNO₃
Then we<u> convert the 150 mL of the volumetric flask into L</u>:
Finally we <u>divide the moles by the volume</u>:
- 0.500 mol AgNO₃ / 0.150 L = 3.33 M
When you say mhello les do uou mean moecules
