\(m_{NaCl}=\dfrac{200\cdot15}{100}=30\left(g\right)\)
\(n_{NaCl}=\dfrac{30}{58.5}=0.51\left(mol\right)\)
\(V_{dd_{NaCl}}=\dfrac{200}{1.1}=181.8\left(ml\right)=0.1818\left(l\right)\)
\(C_{M_{NaCl}}=\dfrac{0.51}{0.1818}=2.8\left(M\right)\)
Ta có: mNaCl = 200.15% = 30 (g)
\(\Rightarrow n_{NaCl}=\dfrac{30}{58,5}=\dfrac{20}{39}\left(mol\right)\)
Mà: V dd NaCl = 200/1,1 = 2000/11 (ml) = 2/11 (l)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{\dfrac{20}{39}}{\dfrac{2}{11}}\approx2,82M\)
Bạn tham khảo nhé!