Câu 1: Giaỉ:
\(m_{NaCl}=\dfrac{200.5}{100}=10\left(g\right)\)
Câu 3: giải:
Ta có: \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)
+) \(V_{ddCuSO_4}=100\left(ml\right)=0,1\left(l\right)\)
=> \(C_{MddCuSO_4}=\dfrac{0,1}{0,1}=1\left(M\right)\)
Câu 3)
nCuSO4=m/M=16/160=0,1 (mol)
Vd d CuSO4 =100ml=0,1(lít)
=> CM=\(\dfrac{n}{V}=\dfrac{0,1}{0,1}=1\left(M\right)\)