Đáp án:

a) 8%

b) 11%

c) 19%

d) 7,4%

Giải thích các bước giải:

a) mNaCl = C% . mdd : 100 = 10 . 160 : 100 = 16g

mdd = 160 + 40 = 200g

C%NaCl = 16 : 200 . 100% = 8%

b) Từ 200g dung dịch NaOH 20%:

mct1 = C% . mdd : 100 =20 . 200 : 100= 40g

 Từ 300g dung dịch NaOH 5%:

mct2 = C% . mdd : 100 = 5 . 300 : 100= 15g

=> mct mới = mct1 + mct2 = 40 + 15 = 55g

mdd mới = 200 + 300 = 500 (g)

C% = mct : mdd . 100% = 55 : 500 . 100 = 11%

c) Từ 360g dung dịch H2SO4 10%:

mH2SO4=C% . mdd : 100=10 . 360 : 100= 36g

mct mới = mct1 + mct2 = 40 + 36 = 76g

mdd mới = 40 + 360 = 400g

C% = mct : mdd . 100% = 76 : 400 . 100 = 19%

d) PTHH: CaO + H2O —> Ca(OH)2

    (mol)  :    1    :    1       :           1

    (mol)  :   0,2 :                        0,2

mdd = 11,2 + 188,8 = 200 (g)

nCaO = 11,2 : 56 = 0,2 (mol)

nCa(OH)2 = nCaO = 0,2 (mol)

mCa(OH)2 = 0,2 . 74 = 14,8 (g)

mdd = mct + mdm = 11,2 + 188,8 = 200 (g)

C% = 14,8 : 200 . 100% = 7,4%

#Creative Team Name

Em tham khảo nha :

\(\begin{array}{l}
a)\\
{m_{NaCl}} = \dfrac{{160 \times 10}}{{100}} = 16g\\
C{\% _{NaCl}} = \dfrac{{16}}{{40 + 160}} \times 100\%  = 8\% \\
b)\\
NaOH\,20\% :\\
{m_{NaOH}} = \dfrac{{200 \times 20}}{{100}} = 40g\\
NaOH\,5\% :\\
{m_{NaOH}} = \dfrac{{300 \times 5}}{{100}} = 15g\\
C\%  = \dfrac{{40 + 15}}{{200 + 300}} \times 100\%  = 11\% \\
c)\\
{H_2}S{O_4}\,10\% :\\
{m_{{H_2}S{O_4}}} = \dfrac{{360 \times 10}}{{100}} = 36g\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{36 + 40}}{{360 + 40}} = 19\% \\
d)\\
CaO + {H_2}O \to Ca{(OH)_2}\\
{n_{CaO}} = \dfrac{{11,2}}{{56}} = 0,2mol\\
{n_{Ca{{(OH)}_2}}} = {n_{CaO}} = 0,2mol\\
{m_{Ca{{(OH)}_2}}} = 0,2 \times 74 = 14,8g\\
C{\% _{Ca{{(OH)}_2}}} = \dfrac{{14,8}}{{11,2 + 188,8}} \times 100\%  = 7,4\% 
\end{array}\)