a) \(n_{NaCl}=\frac{2,925}{58,5}=0,05\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\frac{0,05}{0,15}=0,33\left(M\right)\)
b) \(n_{KCl.0,5M}=0,05\times0,5=0,025\left(mol\right)\)
\(n_{KCl.1M}=0,25\times1=0,25\left(mol\right)\)
\(\Rightarrow n_{KCl}mới=0,025+0,25=0,275\left(mol\right)\)
\(V_{ddKCl}mới=0,05+0,25=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{KCl}}mới=\frac{0,275}{0,3}=0,92\left(M\right)\)
c) Na2O + H2O → 2NaOH
\(n_{Na_2O}=\frac{3,1}{62}=0,05\left(mol\right)\)
Theo pT: \(n_{NaOH}=2n_{Na_2O}=2\times0,05=0,1\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,1\times40=4\left(g\right)\)
Ta có: \(m_{ddNaOH}=3,1+196,9=200\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\frac{4}{200}\times100\%=2\%\)
1)
nNaCl = 0.05 mol
CM NaCl = 0.05/0.15=0.33M
b. nKCl( 0.5M) = 0.025 mol
nKCl ( 1M) = 0.25 * 1 = 0.25 mol
nKCl = 0.275 mol
VddKCl = 0.05 + 0.25 = 0.3 l
CM KCl = 0.275/0.3=0.92 M
c. Na2O + H2O --> 2NaOH
0.05______________0.1
mNaOH = 4g
mdd = 3.1 + 196.9 = 200 g
C% NaOH = 4/200*100= 2%
