Câu 33:
\(m_{NaOH}=500.12\%=60\left(g\right)\\ m_{ddNaOH\left(8\%\right)}=\dfrac{60}{8\%}=750\left(g\right)\\ m_{H_2O\left(thêm\right)}=750-500=250\left(g\right)\\ \Rightarrow D\)
Câu 34:
\(m_{NaOH}=500.20\%=100\left(g\right)\\ m_{ddNaOH\left(16\%\right)}=\dfrac{100}{16\%}=625\left(g\right)\\ m_{H_2O\left(thêm\right)}=625-500=125\left(g\right)\\ \Rightarrow D\)
Câu 36:
Gọi x là số gam KOH nguyên chất thêm vào. (x>0) (g)
\(m_{KOH}=1200.12\%=144\left(g\right)\\ m_{KOH\left(dd20\%\right)}=144+x\left(g\right)\\ m_{ddKOH\left(20\%\right)}=1200+x\\ Có:C\%_{ddKOH\left(sau\right)}=20\%\\ \Leftrightarrow\dfrac{144+x}{1200+x}.100\%=20\%\\ \Rightarrow x=120\left(TM\right)\)
Vậy KL KOH nguyên chất thêm vào là 120 gam
Chọn C
Câu 35:
\(m_{KOH}=200.20\%=40\left(g\right)\\ m_{ddKOH\left(16\%\right)}=\dfrac{40}{16\%}=250\left(g\right)\\ m_{H_2O\left(thêm\right)}=250-200=50\left(g\right)\\ \Rightarrow B\)
Câu 38:
\(m_{1lddH_2SO_4\left(98\%\right)}=1000.1,84=1840\left(g\right)\\ m_{H_2SO_4}=1840.98\%=1803,2\left(g\right)\\ m_{ddH_2SO_4\left(10\%\right)}=\dfrac{1803,2}{10\%}=18032\left(g\right)\\ m_{H_2O\left(thêm\right)}=18032-1840=16192\left(g\right)\\ \Rightarrow V_{H_2O\left(thêm\right)}=\dfrac{16192}{1}=16192\left(ml\right)=16,192\left(l\right)\\ \Rightarrow C\)
