Answered

What is the molarity of sodium chloride solution prepared by dissolving 10.0 grams in 100.0ml of solution?



Answer :

Yipes
[tex]C_{i}=\frac{n_{i}}{V}\\\\ n_{i}=\frac{m}{M_{NaCl}}\\\\ C_{i}=\frac{\frac{m}{M_{NaCl}}}{V}=\frac{m}{M_{NaCl}V}\\\\ m=10g\\ M_{NaCl}=58.5\frac{g}{mol}\\ V=100ml=0.1L\\\\ C_{i}=\frac{10g}{58.8\frac{g}{mol}*0.1L}=1.71\frac{mol}{L}[/tex]

Other Questions