Câu hỏi của Đỗ Nguyễn Khánh Ngọc

Question

Tính khối lượng dung dịch của:

a)20G NaOH 10%

b) 0,1 mol H2SO4 10%

c) 300ml dd HCl ( D = 0,8g/ml)

d) 7,4g Ca(OH)2 10%

e) 0,2 mol HCl 15%

Phùng Hà Châu · Accepted Answer

a) $m_{ddNaOH}=\frac{20}{10\%}=200\left(g\right)$

b) $m_{H_2SO_4}=0,1\times98=9,8\left(g\right)$

$\Rightarrow m_{ddH_2SO_4}=\frac{9,8}{10\%}=98\left(g\right)$

c) $m_{ddH_2SO_4}=300\times0,8=240\left(g\right)$

d) $m_{ddCa\left(OH\right)_2}=\frac{7,4}{10\%}=74\left(g\right)$

e) $m_{HCl}=0,2\times36,5=7,3\left(g\right)$

$\Rightarrow m_{ddHCl}=\frac{7,3}{15\%}=48,67\left(g\right)$Câu trả lời: a) $m_{ddNaOH}=\frac{20}{10\%}=200\left(g\right)$

b) $m_{H_2SO_4}=0,1\times98=9,8\left(g\right)$

$\Rightarrow m_{ddH_2SO_4}=\frac{9,8}{10\%}=98\left(g\right)$

c) $m_{ddH_2SO_4}=300\times0,8=240\left(g\right)$

d) $m_{ddCa\left(OH\right)_2}=\frac{7,4}{10\%}=74\left(g\right)$

e) $m_{HCl}=0,2\times36,5=7,3\left(g\right)$

$\Rightarrow m_{ddHCl}=\frac{7,3}{15\%}=48,67\left(g\right)$

Minh Nhân · Answer

a)

mddNaOH = 20*100/10=200g

b)

mH2SO4 =0.1*98 = 9.8 g

mddH2SO4 = 9.8*100/10=98g

c)

mddHCl = 300*0.8=240g

d)

mddCa(OH)2 = 7.4*100/10=74g

e)

mHCl = 0.2*36.5=7.3g

mddHCl = 7.3*100/15= 48.67gCâu trả lời: a)