Câu 2 :
\(n_{HCl}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(m_{HCl}=0.2\cdot36.5=7.3\left(g\right)\)
\(m_{dd_{HCl}}=7.3+92.7=100\left(g\right)\)
\(C\%_{HCl}=\dfrac{7.3}{100}\cdot100\%=7.3\%\)
Câu 3 :
a)
\(m_{dd_{HNO_3}}=1.25\cdot100=125\left(g\right)\)
\(m_{HNO_3}=40\%\cdot125=50\left(g\right)\)
\(n_{HNO_3}=\dfrac{50}{63}\left(mol\right)\)
b)
\(n_{HCl}=3\cdot1=3\left(mol\right)\)
\(m_{HCl}=3\cdot36.5=109.5\left(g\right)\)
\(m_{dd_{HCl}}=\dfrac{109.5\cdot100}{36.5}=300\left(g\right)\)
\(V_{dd_{HCl}}=\dfrac{300}{1.19}=252.1\left(ml\right)\)
c)
\(V_{dd_{Ba\left(OH\right)_2}}=0.2+1.3=1.5\left(l\right)\)
\(n_{Ba\left(OH\right)_2}=1.5\cdot0.01=0.015\left(mol\right)\)
\(C_{M_{Ba\left(OH\right)_2\left(bđ\right)}}=\dfrac{0.015}{0.2}=0.075\left(M\right)\)
Câu 3 :
d)
$m_{dd\ H_2SO_4} = 150.1,31 = 196,5(gam)$
$m_{H_2SO_4} = 196,5.40\% = 78,6(gam)$
$n_{H_2SO_4} = 78,6 : 98 = 0,8(mol)$
$V_{dd\ H_2SO_4} = 0,8/0,5 = 1,6(lít) = 1600(ml)$
$V_{nước\ cần\ thêm} = 1600 - 150 = 1450(ml)$
e)
$m_{K_2SO_4} = 250.8\% = 20(gam)$
Gọi $m_{K_2SO_4\ cần\ thêm} = a(gam)$
Ta có :
$m_{dd} = 250 + a(gam)$
$m_{K_2SO_4} = 20 + a(gam)$
Suy ra :
$C\%_{K_2SO_4} = \dfrac{20+a}{250 + a}.100\% = 12\%$
$\Rightarrow a = 11,36(gam)$
