1)

a,\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

\(C_{M_{ddNaOH}}=\dfrac{0,2}{0,242}=0,83M\)

\(C\%_{ddNaOH}=\dfrac{8.100\%}{242}=3,3\%\)

b,\(n_{H_2SO_4}=0,1.0,15=0,015\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol:      0,03            0,015

\(C_{M_{ddNaOH}}=\dfrac{0,03}{0,2}=0,15M\)

a)

$m_{H_2O} = 242(gam)$

Suy ra : $m_{dd} = 8 + 242 = 250(gam)$

$C\%_{NaOH} = \dfrac{8}{250}.100\% = 3,2\%$

b)

$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} = 2n_{H_2SO_4} = 0,03(mol)$
$C_{M_{NaOH}} = \dfrac{0,03}{0,2} = 0,15M$

a) \(m_{NaOH}=\dfrac{60.20}{100}=12\left(g\right)\)

\(C\%_{dd.sau.khi.pha}=\dfrac{12}{60+40}.100\%=12\%\)

b) \(C\%_{dd.sau.khi.pha}=\dfrac{12+12}{60+12}.100\%=33,33\%\)

60g dd 20% có 60.20%=12g NaOH, 60-12=48g H2O

a, Pha thêm 40g H2O , ta có 88g H2O

→C%NaOH=12.100:88=12,64% 

b) 

tan thêm 12 g

=>m NaOH=24g

=>C%=\(\dfrac{24}{60+12}100\)=33,33%

\(m_{NaOH}=\dfrac{60.20}{100}=12\left(g\right)\\ a,m_{\text{dd}}=60+40=100\left(g\right)\\ C\%=\dfrac{12}{100}.100\%=12\%\\ b,m_{NaOH}=12+12=24\left(g\right)\\ C\%=\dfrac{24}{60}.100\%=40\%\)

a)

b) nNaOH có trong 200ml dung dịch NaOH 0,25M là

VH2O cần dùng để pha loãng 200ml dung dịch NaOH 0,25M để dung dịch NaOH 0,1M.

VH2O = 500 - 200 = 300ml H2O.

\(n_{MgCO_3}=\dfrac{16,8}{84}=0,2mol\\ a.MgCO_3+H_2SO_4->MgSO_4+H_2O+CO_2\\ 2NaOH+H_2SO_{\text{4 }}->Na_2SO_4+2H_2O\\ b.n_{H_2SO_4dư}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.80.0,1:40=0,1mol\\ n_{H_2SO_4\left(MgCO_3\right)}=0,2mol\\ c.C\%=\dfrac{98.0,3}{200}.100\%=14,7\%\\ V=0,2.22,4=4,48L\\ d.m_{ddsau}=200+16,8-44.0,2+80=288g\\ C\%_{Na_2SO_4}=\dfrac{40.0,1}{288}.100\%=1,39\%\\ C\%_{MgSO_4}=\dfrac{120.0,2}{288}.100\%=8,33\%\)

\(n_{MgCO_3}=\dfrac{16,8}{84}=0,2\left(mol\right)\)

\(n_{NaOH}=\dfrac{80}{40}=2\left(mol\right)\)

PTHH :

\(MgCO_3+H_2SO_4\rightarrow MgSO_4+H_2O+CO_2\uparrow\)

0,2              0,2                                         0,2 

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

2                 1                 1

Vậy có 0,2 mol H2SO4 phản ứng với MgCO3 

       có 1 mol H2SO4 phản ứng với NaOH 

\(m_{H_2SO_4}=1,2.98=117,6\left(g\right)\)

\(c,C\%_{H_2SO_4}=\dfrac{117,6}{200}.100\%=58,8\%\)

\(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

\(d,m_{Na_2SO_4}=1.142=142\left(g\right)\)

\(m_{ddNaOH}=\dfrac{80.100}{10}=800\left(g\right)\)

\(m_{ddH_2SO_4dư}=1.98:58,8\%\approx166,67\left(g\right)\)

\(m_{ddNa_2SO_4}=800+166,67=966,67\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{142}{966,67}.100\%\approx14,69\%\)

\(n_{NaOH}=0,02.2=0,04\left(mol\right)\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ a.n_{H_2SO_4}=n_{Na_2SO_4}=\dfrac{0,04}{2}=0,02\left(mol\right)\\ C_{MddH_2SO_4}=\dfrac{0,02}{0,08}=0,25\left(M\right)\\ b.\left[Na^+\right]=\dfrac{0,02.2}{0,02+0,08}=0,4\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,02+0,08}=0,2\left(M\right)\)

a) \(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

\(C_{M_{NaOH}}=\dfrac{0,2}{0,8}=0,25M\)

b) \(n_{NaOH}=0,2.0,25=0,05\left(mol\right)\)

\(V_{dd}=\dfrac{0,05}{0,1}=0,5\left(l\right)=500\left(ml\right)\)

=> V_H2O = 500 - 200 = 300 (ml)