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%
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