a, mdd= mct+mH2O=10+150=160g

C%=$\frac{mct}{mdd}$ .100%= $\frac{10}{160}$ .100% =6,25%

b, nNH3=8,96.22,4=0,4(mol)

mNH3= 0,4. 17=6,8(g)

mdd= 100+ 6,8=106,8g

C%=$\frac{mct}{mdd}$ .100%= $\frac{6,8}{106,8}$ .100%=6,36%

c,mNaCltrongdd10%=150×10%=15g

mNaCltrongdd20%=200×20%=40g

mNaCl=40+15=55g

mdd=150+200=350g

C%NaCl=$\frac{55}{350}$ .100%=15,71%

CHÚC BẠN HỌC TỐT

Đáp án:

 a) 6,25%

b) 13,11%

c) 15,71%

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
m{\rm{dd}} = 150 + 10 = 160g\\
C\%  = \dfrac{{mct}}{{m{\rm{dd}}}} \times 100\%  = \dfrac{{10}}{{160}} \times 100\%  = 6,25\% \\
b)\\
nN{H_3} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
N{H_3} + {H_2}O \to N{H_4}OH\\
nN{H_4}OH = n{H_3} = 0,4\,mol\\
C\%  = \dfrac{{0,4 \times 35}}{{100 + 0,4 \times 17}} \times 100\%  = 13,11\% \\
c)\\
mNaCl\,trong\,{\rm{dd}}10\%  = 150 \times 10\%  = 15g\\
mNaCl\,trong\,{\rm{dd}}\,20\%  = 200 \times 20\%  = 40g\\
mNaCl = 40 + 15 = 55g\\
m{\rm{dd}} = 150 + 200 = 350g\\
C\% NaCl = \dfrac{{55}}{{350}} \times 100\%  = 15,71\% 
\end{array}\)