a, Ta có : nNaCl = \(C_M.V=1,5.1=1,5\left(mol\right)\)
=> mNaCl = n . M = 1,5 . 87,75(g)
Vậy ...........
b, Ta có : \(n_{MgCl_2}=C_M.V=40.0,05=2\left(mol\right)\)
=>\(m_{MgCl_2}=n.M=2.95=190\left(g\right)\)
Vậy ..........
a)
=>n = 1,5 . 1 = 1,5 mol
=>mNaCl = 1,5 . 58,5 = 87,75 g
- ta có:
m = 50 g => V= 0,05 (l)
=>n = 0,05 . 40 = 2 mol
=>mMgCl2 = 2. 95 = 190 g
Câu a :
\(n_{NaCL}=1,5.1=1,5\left(mol\right)\)
\(m_{NaCl}=1,5.58,5=87,75g\)